 Welcome back to our lecture series, Math 3120, Transition to Advanced Mathematics for Students at Southern Utah University. As usual, be a professor today, Dr. Andrew Missildine. In lecture 33, we're gonna talk about the topic of inverse functions. And before we talk about inverse functions, which of course will continue our topics or our conversation about functions, we actually wanna talk about the more general idea of an inverse relation. So recall, a relation is a collection of ordered pairs from between two sets here. So let R be a relation from set A to set B. So by definition, R is a subset of the Cartesian product, A cross B, okay? Mathematical definition of a relation. So given any relation R, we can define the inverse relation, which will be denoted as R to the superscriptive negative one. You read that as R inverse. R inverse is a relation from B to A. Notice how this direction gets switched around. R is a relation from A to B, but R inverse is a relation from B to A. You're switching the roles of the first set and the second set, and that leads to the relation itself. That R inverse, we think of as a subset of B cross A, like so, and then what it is, it's a collection of ordered pairs, B comma A, such that A comma B was an element of the original relation R. And so in particular, given any order pair, A comma B, the inverse relation switches the orders around. So in particular, if A was related to B with regard to the original relation, this happens if and only if B is related to A with regard to the inverse relation. You just flip that thing around. Likewise, A is not related to B if and only if B is not related to A with regard to the inverse relation. So it just switches those two things around. You swap the order of these things. Now, when it comes to a relation, the order does matter, of course. Now, in the case where A is equal to B, it turns out that a relation is symmetric if and only if it's equal to its inverse relation. All right, so with symmetric relations, we don't have to worry about the order, but in general, the order does matter with the relation and the inverse relation is just switching the order around. So let's see a quick example of this. This is a very simple concept, but still, let's see some examples. So let's take a relation on the set A, which contains one, two, three, four, five, five elements there. So it's a relation from A back into A there. And so let's take the following. So R contains one, one, two, one, two, two, three, three, two, three, excuse me, three, two, three, one, four, four, three, four, two, four, one, five, one, five, two, five, three, five, four, five, five. Okay, so this is a relation we've seen before. This relation, of course, is just the greater than or equal to symbol. One is greater than or equal to one, two is greater than or equal to one, two is greater than or equal to two, et cetera, et cetera. That's all this relation is on the set of five points there. But just thinking of this as just an abstract relation, arbitrary relation, the inverse relation is formed by switching all these ordered pairs around. So one, one becomes one, one. Well, again, when it's the same element, that's just gonna give you the same ordered pair again. But on the other hand, two, one is gonna become one, two in the inverse relation. Two, two, of course, stays the same. Three, three will stay the same. But three, two will turn into two, three. One, three comes from the three, one. Four, four didn't change, of course. Three, four came from the four. Three, four, two will turn into two, four. And then you continue in this pattern, right? Each of these ordered pairs gets switched around and that's exactly where we get this inverse relation. Now, of course, with the original one, I said that this relation is just greater than or equal to on the original set right there. The inverse relation in this setting is gonna be less than or equal to. So one, is less than equal to one, one is less than equal to two, two is less than equal to two, three is less than equal to three. Two is less than equal to three, one is less than equal to three, et cetera, et cetera. So, with regard to these partial orders that we've talked about before, if you have a partial order like in this case, greater than or equal to, its inverse relation is also going to be a partial order. It actually just gives you in the other direction. With regard to equivalence relations, like I mentioned earlier, every symmetric relation is equal to its own inverse. So inverse relations don't really give you anything for equivalence relations, but for partial orders, the inverse relation gives you the counterpart less than versus greater than, or I should say less than or equal to versus greater than equal to as well, okay? But of course, our ultimate goal in this lesson is to talk about inverses of functions. Although I do want to mention that as you draw a relational digraph, which we've drawn this for functions, we've drawn this for partial orders. We can do this for an arbitrary relation, of course. The relational digraph, remember you have the vertices and they're connected to each other by arrows. Well, the inverse relation just takes this exact same graph and just turns all the arrows backwards. That's what the relational digraph will do. So for a partial order, like greater than or equal to, you're gonna take that Hase diagram and flip it upside down because that's gonna switch all the directions around and that's where you get things like less than turns into greater than, et cetera. All right, so now let's get to the main topic here, the idea of a function. What does the inverse relation of a function look like? Because after all, a function is a relation, so it has an inverse relation. Consider a function that we've seen before, take a to be the set, one, two, three, four, b to be the set, one, two, three, four, five, and define the following function relation, one maps to two, two maps to two, three maps to four, and four maps to one, as is illustrated in this diagram right here. One and two both go to two, four goes to one, and three goes to four, exactly like we saw before. Now using the principle we had just learned about on the previous slide, the inverse relation of this function is defined by switching the order of all these arrows. So instead of one going to two, we switch it around to two goes to one. And instead of two going to two, we get that two goes to two. Well, okay, that's the same thing again. But instead of four going to one, we get that one goes to four. And lastly, instead of three going to four, four now goes to three. We switch all of those arrows around. So the graph for the inverse relation, f inverse here, it switches all the directions of the arrows. It's exact same four arrows, but they're now going to opposite directions. In a set theoretic approach, f inverse by definition would be two, one, two, two, four, three, and one, four. And of course, these come from the fact that since f sends one to two, f inverse will send two to one. Two, two, as it's this reflexive ordered pair, it doesn't change at all. But since f sends three to four, f inverse will send four to three. And then since f sends four to one, f inverse will send one to four. So you're reversing the directions here. The input becomes the output and the output becomes the input. But I should mention that while f is a functional relation, it's inverse right here is in fact not a function relation. It's a relation. The inverse relation makes sense is defined, but this is not a function relation. To be a function, remember, every element of the domain has to have exactly one element of the co-domain assigned to it. And so you see some issues here that when you look at the domain of f inverse, if this was a function, the domain is the function, this is the set you're coming out of. So if f inverse was a function, then its domain would now be the set B, which was the co-domain of f. And in some situations, things are just fine. You have one arrow coming out of one. You have one arrow coming out of four. That's exactly what you should see for a function. For a function, there should be exactly one arrow coming out of each element of the domain. So for f inverse, we do fine with one and four, but of course there's some issues here. The issue, for example, comes with two. You'll notice there are two arrows coming out of two here, which is in violation of the vertical line test. That is a function cannot have more than one element assigned to it. So we have the order pairs two, one and two, two. So what is f inverse of two? Is it one or is it two? Right, we don't know. And therefore, this function is not well defined. It's not a function, okay? But we also have issues with like five and three, for which there's no arrows coming out of it whatsoever. Now I should mention that the fact that there's two arrows coming out of two is really an observation that the function f is not one to one. Because if it was one to one, then there would be no points in the co-domain that have more than one arrow pointing to it. So because the function f was not one to one, its inverse has multiple arrows coming out of it from a single location. Also, because the function is not surjective, it's not onto, when you take the inverse relation, you're gonna get points in the new domain that don't have any arrows coming out of it. So if we want the inverse of a function to likewise be a function, it turns out that it needs to be one to one and it needs to be onto. That is, it needs to be a bijective function. And so let me give you an example of such a thing here. Let's switch things up a little bit. Let's consider this time the sets one, two, three, that's a, and b will be the set a, b, c. I mean, I could have called it one, two, three again, but we'll just switch it around just to avoid any confusion, just give some variety here. And so take our function f this time, f sends one to c, it sends two to a and it sends three to b. That's what it does. So one goes to c, two goes to a and three goes to b. So now if we were to change the arrowheads on each and every one of these and draw them on the other side, that's exactly what the inverse relation is gonna do. We can see that illustrated right here. So now a goes to two, b goes to three and c goes to one. So we switch all of these order pairs around. One goes to c for f, so f inverse will send c to one. For f, f sends two to a, therefore f inverse will send a to two. And lastly, since f sends three to b, f inverse will send b back to three. And so you just switch the directions of all these arrows here. And so now when you look at f inverse, this is in fact a function. Every element of its domain has exactly one arrow attached to it, one and only one. And this is exactly a result of the fact that our function was a bijection in the first place, that this was one to one. You'll notice that every element in the co-domain has at most one arrow pointing to it. So it's one to one. It's also surjective, it's onto because every point in the co-domain has at least one arrow pointing to it. And that makes it a bijective. So there's this so to call one to one correspondence, one to one correspondence between the elements of the domain and the elements of the co-domain. One to one correspondence is often used as a synonym for a bijection. And so we do see that when our function wasn't bijective, the inverse relation was not a function. And we see in this example that when the function is a bijection, then its inverse relation is in fact a function. And that is in fact always the case. Now, before we prove such a statement, let's actually introduce a slightly alternative definition of the inverse function, not using relation. So we're gonna see that the two things coincide with each other. Now, in order to do that, I wanna first introduce a very important but also very simple function. Consider the function F, which maps between the set A to A. So it's the same set. The domain and co-domain are the same thing. And this function is defined by the rule that F of A equals A. That is this function always maps the element A back to itself. This is referred to as the identity function. And it's typically denoted as ID sub A. You can drop the A when the function is, when the domain of the function is obvious. But typically if we need to reference multiple identity functions at the same time, we put a little subscript to represent the identity on what set, okay? And I should mention that the identity function is always a bijective function. Okay, it's always bijective. It's always one to one. It's always on two. And this is very easy to see by this formula here. So like if you have that F of A equals F of B, well, F of A maps to A and F of B maps to B. And so connecting these things together, we see that oh, A equals B. This is a one-to-one map. Why is it on two? Well, if A is inside the co-domain A, then we have that F of A is gonna map onto A. So it's on two. So we can very quickly see that the identity function is bijective. And now we wanna introduce this for the following definition. But one other thing I do wanna mention here, this is a very nice result that I'll leave it to the viewer to prove here. So we can write this as a theorem. So you have a relation. So a relation here, a relation, R is a relation on A. Then the following fact is very, very simple to prove here. R is in fact reflexive, if and only if the identity function, which is itself a relation, can be viewed as a subset of R. Very simple fact. I'll leave it to the viewer to prove such a thing. All right, so now we're turning to our conversation about inverse functions. That is the titular topic after all. So let's define an inverse function and actually we're gonna prove that this definition of inverse function is actually equivalent to the inverse relations we were talking about in the realm of bijective functions. So give us a moment to develop that theory. So imagine we have a function F from domain A to codomain B. We say that F is invertible if there exists a function F inverse from B to A and this function is called the inverse of F that satisfies the following condition. That if you compose together the functions F inverse with F, this will give you the identity on A and if you compose F with F inverse, this will give you the identity B. So the inverse function has the property that when you compose it, when you compose a function with its inverse, you always get back the identity function where it's the identity on either the domain or codomain depending on which one you're doing. So a function is invertible if it has an inverse and the inverse is exactly the function which composes to give you the identity in both directions. So let's look at some functions we might have seen before like in a calculus setting and see those, what their inverse functions would look like. So for example, take this function F which is a function from the real numbers to the real numbers. It's defined by the relation F of X equals X cubed. Some of you probably can guess what's going on here. The inverse function does in fact exist. The inverse function will be a map from R to R and it's actually defined by the relation F inverse of X is going to be the cube root of X like so this function, the cube root of X it has as its domain all real numbers it outputs all real numbers and you can very quickly see that if you compose these things together so F inverse, F of F inverse here you're gonna take the cube root of X cubed which gives you back X. Likewise, if you flip this thing around F inverse of F here of X you're gonna get the cube root of X cubed this likewise gives you back X. This function does have an inverse the inverse of X cubed is gonna be X the cube root of X as well and I do wanna mention that this function is in fact a bijective function. We have seen this before and it's inverse is not a quinketing we'll actually see in the future that that's exactly the case. Invertible functions are exactly bijective functions. All right, now let's consider the function G here which is a function from all real numbers to all positive real numbers given by the function G of X equals E to the X. Now note here, the reason we take the co-domain to be positive real numbers is that in order for this map to be surjective I need it to be the positive real numbers. E to the X does not map onto all real numbers you only can get a positive number E to the X is always a positive number there and so in order for this thing to be bijective I do need the co-domain to be only the positive numbers there. I need the co-domain in the range to agree with each other. This function does have an inverse, right? Again, we've probably seen this before. The inverse function in this case G inverse is gonna be a map from zero to infinity so all positive real numbers, it maps to the real numbers and it's given by the formula that G inverse of X is equal to the natural log of X and sure enough that when you compose these things together G of G inverse of X you end up with E to the natural log of X power there. Now the natural log of X is exactly the power of E that gives you X so this composes and becomes an X. If you flip this thing around G inverse of G of X here you're gonna end up with the natural log of E to the X, right? And so then you're asking yourself what power of E gives you E to the X? Well, that's X right there. These are exactly inverse functions. Another curious thing I should mention about inverse functions which we probably saw this before is that you switch the roles. You switch the roles of domain and co-domain in these situations. Now we will see in just a second that for a function to be invertible it does have to be bijective. So the domain, excuse me, the co-domain and the range are identical for a bijective function. The inverse function switches the roles of domain and range in that situation. For an inverse function the domain of the original function becomes its co-domain becomes its range and because it's likewise gonna be a bijective function. The inverse function is itself bijective because the inverse of a function is itself invertible because it has an inverse. G inverse has an inverse because G is its inverse, right? So this gives you a very simple property that G inverse inverse is actually the original function again, okay? So the inverse, the G inverse will take the domain and make it its range and likewise it'll take the co-domain, its range and turn it into its domain. It swaps the roles. Input becomes output and output becomes input. But one needs to be very careful when you consider inverse functions because of examples like the following. Take H1 here to be the function from all real numbers to all non-negative real numbers. So zero to infinity inclusive on zero there defined by the formula X squared, okay? Now when we look at this function right here it's very tempting to say it's invertible because we're like, oh yeah, H inverse which would have to be a function from zero to infinity to the real numbers because again you swap the domain and co-domain there. This will be defined by the rule H1 inverse of X is equal to the square root of X, okay? It's very tempting to say this is the inverse function but the problem is the following. If I compose these things together if I take the square root of X and I square that that does give you back X for which you could be like, oh, that's the identity on X right there. The other way around though doesn't give you the same thing. If you take the square root of X squared this does not give you X. This actually gives you the absolute value of X which is not the same thing as the identity of X because if I take a negative number I'm actually gonna switch it. For example, if I take like two squared and take the square root, well if you follow through with that calculation you get the square root of four which gives you back a two. So sure that act like the identity on two but if I take something like the square root of negative two squared negative two squared is again positive four which gives you back two. And so this is actually evidence that our function X or H one here is not one to one. It's not an injective function because actually what happens is that the square of two and the square of negative two gives you the same thing and then the square root of X gives you back the same thing in that situation. In particular, this is not negative two. So this is why it's necessary to check the composition in both directions. While it looks like we have inverses in one direction we didn't in the other direction because the function was not one to one. And so we can actually repair this thing here. So we'll take our second attempt. We'll take H two here for which H two since our function's not one to one we can shrink the domain to actually be zero to infinity there and maybe I'll just write this down below it's getting too messy. If we define H two to be the function from non-negative reals to non-negative reals using the exact same formula so H of two is still equal to X squared. Then in this situation our function is now one to one because we threw out the negative numbers which cause repetition with regard to the squaring function and it is onto every real number can be written as a square of a real number every non-negative number there negatives can't do that because you need to imagine numbers to do that. So if we restrict the domain we do now have a bijective function and sure enough H two here does have an inverse. You switch around the domain and co-domain looks the same thing because the domain and co-domain were the same there and then H two inverse of X that is the square root of X. So the thing is X squared is only an invertible function if you choose the right domain and co-domain and this is a subtle point that often is not considered with full respect in like a calculus or pre-calculus setting because the set theoretic notions of domain and co-domain don't get their proper treatment in that situation in that the pre-calculus setting functions are thought of as formulas not as these maps, not as these relations and as such you like oh well the formula X squared has an inverse function but that's not that that leads to confusion and errors on the students part because of their incomplete understanding of what a function is. Hopefully as we transition into advanced mathematics we will not make that mistake that many pre-calculus students make as well. Now one other thing I do wanna make mention about these inverse functions is that this definition of the inverse of an invertible function that is a function's invertible if it has an inverse I remember that the definition was up here, right? We say that a function is invertible if it has an inverse, okay? This definition is equivalent to what we talked about earlier about the relational inverse that is if an inverse exists it has to be the relational inverse that we were talking about beforehand because this idea of composition when you take a relation here that you have some point because again you think of this as a point, right? Let's say that the function F it contains the element A comma B, okay? Then if you compose that, let me write it up here. So if our function F sends A to B that means it contains the ordered pair A comma B. So this statement here is equivalent to the statement that F of A is equal to B. Now function composition then says that if I compose this with F inverse so if you take F inverse of F of A you're gonna get F inverse of B here this statement right here says that F inverse of B has to equal A. And so this statement then translates the same that the ordered pair B comma A is inside of F inverse. So what we get here is that A comma B being inside of F implies that B comma A is inside of F inverse. So if the inverse function exists it actually has to be the inverse relation which is why we're justified in calling these things by the exact same symbol here. But what about this business about bijectivity? I can only kick that cane down the road for so long. So let's now consider that statement here. So let's prove the main topic for this video here. A mapping is invertible. The function is invertible if and only if it is one to one and onto. That is a function has an inverse exactly when it is a bijection. The inverse relation of a function is itself a function only when the function is injective and surjective. Now this is an if and only if statement so therefore there's two directions to go. We have to first assume that the function is invertible then show it's bijective and then conversely we have to show if the function is bijective then it has an inverse. There's two different notions there. We have to prove both of them to be the case, okay? And it doesn't actually matter which direction you can go. I'm actually gonna start with this one. So let F be our function. F is a function from A to B here. That is the mapping we're gonna talk about. And so let's suppose that F is bijective. It is one to one and onto. Now since F is bijective then for each B in the co-domain there exists a unique A inside of the domain such that F of A equals B. Notice in this statement right here I'm using simultaneously that it's one to one and onto. So the fact that A exists, there is an A that's using the surjectivity. Because it's surjective given any B there is an A inside of the domain that maps onto B. That gives us the existence of this element A. But I also know that the element A has to be unique because if there was a second element that does it then it wouldn't be one to one. And so the bijectivity assumption gives us there exists a unique element of the domain which means unique means it exists but it's also only one. There's a unique element such that F of A equals B there. And since this is a unique element we can actually define a function using that. We define a function G which is a relation from B to A and the relation is exactly this. We get that G of B is equal to A. We define a function relation there exactly when F of A equals B. Now because of the previous assumption about bijectivity given any B there is always a unique A. And so the uniqueness statement mentioned above means that this relation G is in fact well defined. There is only one order pair associated to the element B. There's only one element in A that mapped onto B via the function F and therefore G is making that connection. It's reversing it around, okay? So then when you compose these functions together so we do get the uniqueness statement gives that G is a well-defined function. It's a function. And when you compose these together F of G is equal to the identity on B and likewise G of F is the identity on A. And this is very straightforward to see here because if I do the second one first, right? If you take G of F of A, well F of A is gonna map to some B but by definition G when you take an element B maps to the unique element that mapped onto it by F and that's gonna be A, all right? And then the other direction and if I take F of G of B, by definition G will map B to the element of A which we'll call little A that maps onto B via F and therefore F will send that thing back to B. So by construction, these things are inverses. I mean what we constructed here, this right here is the inverse relation of G. Sorry, G is the inverse relation of F by the construction. We just switched around all the order pairs. That's exactly what these statements are saying here. So we're saying that when A comma B is inside of F then we say that B comma A is inside of G which of course makes it the inverse, okay? But because of the bijectivity, this relation will in fact be well-defined. It gives us a function. Now let's go the other direction. If F is an invertible function, that means it has an inverse. That means there exists some function F inverse such that F composed with F inverse is equal to the identity on B there for which I should mention that the identity, like we mentioned before, is a bijective function. So in particular, this function is one to one and onto. For this situation, I wanna note that the identity is an onto function. Now in our homework, right? We didn't do this in the lecture series here but in our homework we have theorem 32.1 which actually says that when you compose two functions together, if the composition of two functions is surjective, then the left function is likewise surjective. So theorem 31, 32.1 tells us that since this function is onto, this function will be onto as well. So we see that F is an onto function. Now likewise, because the function's invertible, you can switch the directions and we get that F inverse of F is likewise equal to the identity. A different identity function potentially. It's identity on A now but nonetheless, this function is still bijective and this time I wanna note that the identity function is an one to one function. Now if we use theorem 32.2, which again, this is a theorem left as an exercise to my students here, the statement of 32.2 tells us the other relation here. If a composition of functions is equal to an injective function, that means the right function is likewise injective. So using theorem 32.2, we get that F is likewise injective. So since F is onto and it's one to one, that means F is a bijection and that then proves the main result here that we have that a bijective function is exactly an invertible function. The two notions are one of the same thing. I wanted to conclude this video with an example coming from linear algebra here. Consider A to be an M by N matrix. So it has M rows and N columns here. Then this matrix naturally produces what we call a linear transformation. This is a map from the vector spaces Rn to Rm and it follows the following rule here that the transformation takes a vector X and it transforms it into the product of A times X. So this is a transformation in linear algebra. They often call them transformations. This is of course is a function using our terminology here. And so yeah, this is a function. Linear transformations are very important functions in linear algebra. And as such, I wanna say a comment about these things here. As we've learned in linear algebra class, like math 2270 using Southern Utah University's course number in here, an M by N matrix. So one that has the same number of rows as columns. So it's a square matrix sometimes called. So an N by N matrix is in fact invertible if it has an inverse, right? So that means there exists some inverse matrix A inverse such that you get that A times A inverse is equal to A inverse times A, which is equal to the identity matrix in that situation. So that's what it means for a matrix to be invertible. Now it turns out that if you have a linear transformation created by a matrix, if it's invertible as a matrix, then the inverse matrix produces the inverse function. So a linear transformation is invertible if and only if it's corresponding matrix representation is invertible as well. So you can connect this invertibility of functions with invertibility of matrices. But what about the bijectivity there? When it comes to this matrix equation here, you're often interested in things like the following. You consider the equation AX equals B. Well, this system is always consistent. Like this problem is always consistent, meaning that, oh, given any B, given any B this equation has a solution, that would happen if and only if the function T is in fact onto, okay? So the fact you could always get a solution is because this is onto linear transformation. And likewise, do you have multiple solutions or do you have a unique solution to this equation, right? Are there more than one X that can solve this equation? That statement is exactly equivalent to saying that T is in fact one to one. So AX equals B has a unique solution when T is one to one and AX equals B has at least one solution when T is onto because it doesn't matter what you choose for B or X, you can always produce those things, okay? Now, the reason I wanted to make mention of these things is because these conditions together give us a so-called non-singular matrix. So using matrix vocabulary, a matrix is non-singular if and only if its corresponding transformation is bijective, okay? So again, as you take properties of matrices and compare it to the corresponding linear transformation, a matrix is non-singular if and only if it's bijective. But then connecting this to what we see before that a matrix will be non-singular if and only if it's in fact invertible. And so I wanted to bring up this example because this is a special case of the theorem we just proved. In linear algebra, a matrix is invertible if and only if it's non-singular, which its corresponding linear transformation will be invertible if and only if it's bijective. But this in fact holds in general, not just in this linear algebra case, but this is a fun case to review because linear algebra is pretty awesome, all right? Now with that, we come to now to the end of lecture 33. Thanks for watching. If you learned anything about inverse relations or inverse functions, please like this video, subscribe to the channel to see more videos like this in the future. If you liked this video or learned something from it, share with friends and colleagues so they can benefit from it as well. And as always, if you have any questions, feel free to post them in the comments below and I'll gladly answer them as soon as I can.