 In this video, we're going to prove that a function is invertible, if and only if it's bijective, that is, a function's invertible exactly when it's one to one and on to. And you might have noticed, if you saw the previous video here, that if a, like the square, the squaring function, right, we had h of x equals x squared. What we saw there is if we took this, if we took the domain to be r to r, we got problems. We didn't have an inverse function. But when we restricted the domain to be zero to infinity, that actually makes this function one to one. And when we restricted its co-domain to be zero to infinity, that makes the function on to. And therefore, by restricting the domain and co-domain to have an acceptable inverse function, we're forcing the function to be bijective like we saw before. This isn't a coincidence. This is actually necessary in order to have an invertible function. So this proof right here is our standard if and only if type proof. That is, we're trying to show a bi-condition that two things are logically equivalent to each other. So this actually represents two proofs for the price of one. So let's say that the function being invertible is property a, and the function being bijective is property b. In order to show that a is equivalent to b, we have to show, so we're basically showing the following. If a logically implies b, and if b logically implies a here, then in that situation, we're going to get that a is equivalent to b. So to show an if and only if statement, you have to show two conditionals. You have to show that sufficiency implies necessity, and necessity implies sufficiency, the two directions we could go here. And it honestly doesn't matter which one you start with, just pick the one you want to. Oftentimes, I'll just follow the order they were listed, right? And so you could be like, if our function f is invertible, then show it's bijective or vice versa. In fact, the way that this proof is written, I actually started with, we assume that the function is bijective. And we're going to use that to show that the function is invertible. Because there's two different meanings of these things, bijective means one thing, invertible means something else. So if a function, let's start with that. So we're going to let f be our function, and it'll go from a to b, it's domain and codomain. And so let's assume it's bijective, bijective meaning one to one and onto, right? And so that's what it means to be bijective. So what we're doing here is if we're going to make an assumption, right? So our function is bijective, then we have to show that it's invertible. What does invertible mean? It means that there exists some f inverse that goes from b in order to a. And it must be true that when you compose f with its inverse, you're going to get the identity map in either direction you go. That's what we're trying to show. So we assume bijectivity and we're going to then construct an inverse function from that assumption. So we assume f is bijective, which means we're one to one and onto. And so we now have to define a function to be the inverse. So we're going to define a function g by the following rule. Well, since the function is bijective, for each b in the codomain, there exists a unique a such that f of a equals b. And this is unique, right? Because it's surjective for every b, there is at least one a that maps onto it. And because the functions one to one, no two different a's will map to the same b. So there is this one to one correspondence between the values in the domain and the values in the codomain. That's what we get from being bijective. So we're going to define a new function g, which will be a map from b to a. And we define it by reversing this relationship. If f sends a to b, then g is going to send b to a we just reverse that around. And since we have this one to one correspondence, this map right here is going to be well defined. That would be a concern that one might have. But there's no problem with the definition of this thing because f is a bijective function. All right, then one has to prove that this thing is that when you compose these things, you get the identity, right? If you take f of g evaluated at b, the domain of g is b, right? Well, g will map, so this is just f of g of b. Now g will map b to the unique element in a that maps to b via f. So we get f of a. Now, by construction, a is the number which if you plug into f is going to give back b. So you can see that f of g, f of g evaluated at b is equal to b. And vice versa, if you take g of f evaluated at a, this will be g of f of a, which f of a is going to map to some number b. But g is the function which if given b, it will map you back to a, so you get back a. So in both situations, so you get from b to b, that's the identity map. You go from a to a, which is the identity map. And so as b and a were chosen arbitrarily, what we then see is that f of g is just equal to the identity of b. And g of f is just the identity at a. So this shows us that in fact g. So what this then gives us is that g is an inverse of f and there's only one inverse. So g is an inverse of f. Therefore, f is invertible. This gives us the first direction. This shows that this shows that bijective implies invertible. We want to go the other way around now. We want to show that invertible implies bijective, which means we want to show it's one to one. And we want to show it's onto. I'm going to clear out this space a little bit. So I have some room to write. So what we're going to do now is let's assume the function is invertible. If it's invertible, it has an inverse. So there exists some function f composed with f inverse that gives you the identity. All right. Now the identity map is, well, it's the identity map we mentioned in the previous video that the identity map is in fact a bijective map. It's both one to one and onto. And so by a homework question, which you can see in the textbook, it's assigned to you by exercise 1322 part B in Judson's textbook. This shows us that if a composition of two functions is surjective, which the identity is, then this will show you that the left factor in terms of the composition is likewise surjective. So that gives us that f is surjective because it can compose to a bijective function. Now using that same exercise, let's go the other way around. Let's take f inverse of f this time. This composes to be the identity. The identity, again, is a bijective map. So in particular, it's one to one using that same exercise. But this time part C, part C says if a composition equals an objective map, then the right factor must have been injective. Again, my students will have an opportunity to work through this one in their homework. And so because f has an inverse, it's both one one and onto, therefore it has to be a bijective function. And this thus finishes the proof of this one. Invertibility of a function is equivalent to bijectivity of that function. And for many of us, we've kind of seen this experience before. So let me borrow an example from linear algebra here. Let A be an M by N matrix. So it has M mini rows and N mini columns. Rows then columns here. So given any matrix A, one can naturally produce a linear transformation from RN to RM. Remember, N is the number of columns, M is the number of rows. If you take a vector in RN and you times it by an M by N matrix, you'll end up with a vector in RM. So this is a vector in RM. RM, of course, is the arrays of vectors with M entries, like so. So associated to every matrix, there's a natural function which we can ascribe from that matrix, right? This is a very standard construction in linear algebra. As one would probably have learned in linear algebra class, like at SUU, this is Math 2270, for which actually I have many videos about that. Feel free to check out that channel. If you have an N by N matrix, that's sort of like a special situation right there. If you have an N by N matrix, then we know that a matrix is invertible if and only if, well, we know a matrix is invertible. It has an inverse, A inverse. This will be happening if and only if the matrix is non-singular. Now, non-singularity can mean a lot of things. One way of interpreting it is non-singularity would imply that the equation AX equals B has a unique solution. So you have a unique solution for any choice of B, right? So notice that this statement has a special case of the previous theorem. We know that A being non-singular is the same thing as it being bijective, because the uniqueness of solution right here comes from the following idea. The system will be consistent only if AX equals B for some choice of X, right? In terms of the linear transformation T, that would imply that T is a one-to-one map if this always has A solutions, always consistent. And the fact that it has a unique solution, this comes from the linear independence of the columns of A, which can be connected to the injectivity of this map right here. This map T, this transformation will only be one-to-one when A has linearly dependent columns, which will mean that this equation has at most one solution, all right? So the uniqueness, the non-singularity definition of this uniqueness of solution is equivalent to the linear transformation being bijective. So note that AX equals B has a solution. This means that B is in the image of the map T, so there's something there. And again, uniqueness means it's one-to-one. And so these notions, this proposition, or did I call it a theorem? I don't remember now. Theorem 1225, this is a generalization of that principle we saw, this non-singularity theorem one sees in linear algebra. Because this is true in general for functions. We'll have an inverse if and only if the function is bijective, a very important principle to remember.