 This lecture is part of an online course on homological algebra and will be about basic properties of the group Tor A and B for A and B abelian groups. So the properties we want to cover as follows. First of all, we need to check that it is well defined. So in the previous lecture we define Tor A and B by choosing a resolution not goes to Z to the M goes to Z to the N goes to A goes to naught and then tensioning this with B. So we get not goes to Tor A B goes to B to the M goes to B to the N and it's not clear that this is independent of the choice of resolution of A. So secondly, we want to show that Tor A and B is a functor in A and B. What this means is that if we've got a map, a homomorphism from B to C, then this induces a natural homomorphism from Tor A B to Tor A C, which behaves nicely with all the usual categorical properties. Third basic property is that Tor is symmetric in the sense there's a natural isomorphism between these two groups. So we calculated Tor for finite abelian groups and we noted that this holds for finite abelian groups and now we want to check it for all groups. Finally, we want to check the long exact sequence. So this says that if we've got an exact sequence, naught goes to A goes to B goes to C goes to zero, then we get the following long exact sequence. We get Tor A to M goes to Tor B to M goes to Tor C to M goes to A tensor M goes to B tensor M goes to C tensor M. So the Tor groups sort of control the lack of exactness of the tensor product. So we'll start with trying to show that Tor is well defined. So suppose we've got, so to define Tor, we need to take a resolution of A and we might take a second resolution of A and we want to show that Tor calculated using these two resolutions is actually the same. And usually you denote the, when you've got a resolution like this, let's denote this map by D and we're rather overusing D for every possible map. D comes from the boundary map in topology, of course. And to start off with, we want to compare these two sequences and it's obvious how to compare A with A because we've just got an isomorphism there. And now what we can do is we can find a map from this Z to the N1 to Z to the N2, making this diagram commute. And this exists because Z to the N1 is free. So for each basis element, we just map it to something whose image is the same as that. And now we can continue this trick and find the map from Z to the N1 to Z to the N2, making everything commute. So we can extend the identity map from A to itself to a map from this resolution of A to the other resolution of B. So this means that if we take, if we tensor with B, we get B to the M1 goes to B to the M2 goes to A and we've got B to the M2 goes to, so it should be a N1 B, B to the N2 goes to A. And we've got maps here and here. And this induces a map from Tor A, B. So let's call the kernel of that Tor A and B. And now we've got a different Tor A and B. Let's call this Tor Prime A and B. So what we want to know is that these two groups here are isomorphic. And this map from between these two groups obviously induces a map between these two groups. So at least we've got a map from this group to this group. However, we now render into this problem that these maps here are not unique. So it's not at all clear that we've got a unique map like this. These two maps aren't unique. We show this, the induced map here is unique. So how can we show this is unique? Well, what we do is suppose we've got two different maps defined like this. So suppose I've got a green map and a pink map. What I'm going to do is I can define a new map. Let's have a different color so I can see what it is. I guess that's almost the same as black. And I'm going to call this map S. And S is going to have the following property. Let's call the green map F and the pink one G. And I'm going to choose S. So if we start here and first do S then D, then this is equal to F minus G. So this identity here is basically talking about this triangle here. It says that if we go around like that, it's the same as the difference between these two maps. And again, S exists because if we take F minus G and apply D to it, this is zero, which means the difference of F and G must be in the image of this map here. So we can lift this map here. Okay, now we take a look at what happens on this other triangle. I guess I'm running out of colors a bit here. So what I want to do now is look at this triangle here. And we notice that here we have the identity that F minus G is equal to SD. And this isn't a misprint. It really is the other way around. So here we first do D, then we do S. So here we have a DS and here we've got an SD. And this follows quite easily because we notice that D F minus G is equal to F minus G D, which is equal to D SD. So D of this map is the same as D of this map. So since D is injected, they must be the same. By the way, we can combine these two identities by saying F minus G is equal to SD plus DS. That's if we extend S to be zero here and here. And this condition here is very important. It turns up all the time in homological algebra. And we say F and G are homotopic. This is very closely related to the condition in algebraic topology that two maps are homotopic. Anyway, we can now use this to show that these two maps are isomorphic. Sorry, these two groups are isomorphic. So what we notice is that if we've got two maps, so we've got tra A to B, goes to B to the M1, goes to B to the M1, goes to 12 prime AB, goes to B to the M2, goes to B to the M2. We notice that if we've got a map here, we can still call this F. So we're taking this map here and tension it with B and we still might have two maps. We still might have some maps F. Let's use colors so I can actually see which is which and we might have another map G. And these still satisfy the same identities as before. So we know F minus G is equal to S times D where S is some map going like that. So what this says is that F is so this says that F and G are the same on the kernel of D. So if D applied to something equals naught, then F must equal G. And these two, so this map is in the kernel of D. So F and G must actually be the same here. So here we have F is actually equal to this map G. So the fact that F and G are homotopic on this resolution means they're actually the same on the tour groups. So what this means is we get a well-defined map from tour AB to the other tour AB, where this is defined using the first resolution of A and this is defined using the second resolution of A. Well, obviously we can do exactly the same thing with the second resolution of A and we get a well-defined map backwards. So we get really two well-defined maps between these. And now we notice that the composition of these two maps is the canonical map from tour A to tour A induced by the identity resolution, by the identity map on the resolution of A. So the composition of these two maps is the identity. And similarly, the composition in the other direction is the identity. So these two groups are isomorphic. And then not only isomorphic, but there's a canonical isomorphism between them. So this shows that tour really is well-defined and independent of the resolution because for any two resolutions, there's a canonical isomorphism between the two tours. Now we can check that tour is actually a functor and this is fairly easy. So it's a functor in A and B. We're taking tour of A and B. And to see this, if we've got a map from B to C, what we want to do is to show there's a map from tour AB to tour AC. And for this, we just take any old resolution of A and we tense it with B and C. So we get naught goes, well, to, sorry, it's not a naught there. We get B to the M goes to C to the, goes to B to the N. And we get maps from C to the M goes to C to the N. And the kernels of these maps are the tour groups. And obviously, if we've got maps like this, making this commute, this induces a map from tour AB to tour AC. And you can check, as before, this is independent of the choice of resolutions. So we get a well-defined, so if we've got a map from B to C, we get a well-defined map between the corresponding tour groups. And you can check this makes tour into a functor in its second argument. We could also prove that it's a function of the first argument, but since we're about to show it symmetric, we don't really need to do that. It will follow immediately from the symmetry. Next, we want to show that tour AB is naturally isomorphic to tour BA. And the definition isn't symmetric because we start by choosing a resolution of A and tension it with B, whereas here we start with a resolution of B and tense with A. And it's not at all obvious that they're the same. So let's take a resolution of A and let's take a resolution of B. And how do we compare these tours? Well, the idea is we tense these resolutions together. That's getting a huge diagram, which I'll now spend the next few minutes writing out. First of all, we tensor the resolution of A with Z to the S. And then we tensor it with Z to the T. And there are maps between these groups. And then we tensor it with B. So we get Z to the M tensor B goes to Z to the N tensor B goes to A tensor B goes to zero. And these maps all onto and you notice here, since we're tension with Z to the N, it preserves exactness. So we get zero there and we get zero there. And if you look at this diagram, you see I've actually forgotten to push in a zero there and a zero there. Well, I didn't forget because these maps aren't injected. And what we get here is tour A B as the kernel. And up here we get naught goes to tour B A because we've taken resolution of B and tense it with A. So what we want to do is we've got this huge diagram of far too many exact sequences. And we want to show that this group at one end of the diagram is isomorphic to the group at the other end of the diagram. So how do we get from one group to the other? Well, we get it by doing a sort of zigzag. So we start here, then we take the image in here. Then this map is onto so we can lift the element up to here and then we can take its image in here. And then the image here is zero because it's the image of this. So we can lift it up to an element here and then we can take the image here and that finally gives us an element there. So we see we've got this sort of zigzag. And this gives us a homomorphism from this group to this group or does it because there's a bit of a problem here. But it's not all clear this map is well-defined because when we lifted this element here, it wasn't unique. So we don't actually get a homomorphism from this group to this group because there's an ambiguity. We could change how we lift the element, but if we change it, it differs by an element here. And this doesn't really make any difference because the image of an element here in this group is zero. So although the map from here to here isn't well-defined and the map to there isn't well-defined either, the map from here to here is well-defined. So we get a homomorphism from tour A B to tour B A. And you can do exactly the same thing the other way around. We get a homomorphism back the other way and you can see that these homomorphisms are inverses of each other. So this isn't just an homomorphism. It's an isomorphism. So that's why tour is symmetric. It depends on this wonderful zigzag. Final thing we want to check about tour is that it satisfies a nice long exact sequence. So if we've got a map naught goes to A, goes to B, goes to C, goes to zero. We want an exact sequence involving tour and the tensor product, which I'm not going to write out yet again. And it's not quite yet. So how do we get this? Well, we want to tensor this sequence with M. So let's take a resolution of M naught goes to Z to the M, goes to Z to the N, goes to M, goes to zero. And now we're going to tensor this sequence with A. So we get A to the M, goes to A to the N, goes to A, tends to M, goes to zero. And the kernel of this is tour A, M. And then we're going to tension it with B. So we get naught goes to tour B, M, goes to B to the M, goes to B to the N, goes to B, tends to M, goes to zero. And that will come as no surprise. We now do the same to C. So we get tour C to M, goes to C to the M, goes to C to the N, goes to C tends to M, I guess, goes to zero. And now we can fill in all these. A lot of these bits are exact. This row is exact and this row is exact because we're just tensioning with a free group. And we get induced maps like this and this map here is obviously injected because they're submodules of this and this. And now there are two missing zeros. We haven't put a zero there and we haven't put a zero there. And that's because this map isn't usually subjective and this map isn't usually injective. And what we want is to define a map that goes all the way from this corner down to this corner of the diagram. So this is sometimes called a snake map for fairly obvious reasons. It sort of looks like a snake winding its way through there. What we want to do is to show that if we take this sequence, if you go along here, along here, and along here, then this orange line is exact. And so how do we do this? Well, we want to define a map from this group to this group and we do it by using a zigzag. So we're going to start here and then we can take the image in here and then this map here is onto so we can lift it like that. And then we can take the image in here and then we can, since the image there is zero because it's the image of this, we can lift it to something in here and then take the image here. So we've got a zigzag in a, well, it's not quite the same sort of zigzag as we had before because it's a different diagram, but we've got exactly the same problem. We want to show this map is actually well defined. And again, there's this problem that the lift from here to here is ambiguous because we could take two different elements with the same image there. But if you took two different elements like that, the difference would be in the image of this element here. So when we go down to there and back to there, the image is well defined because it's just the, sorry, the image in here is well defined because it's just the image of this element here whose image in there is zero. So if we start with an element there, its image there isn't well defined, it's only well defined up to the image of something there. Its image there isn't well defined, it's only well defined up to something there, but the image here is well defined. So we get a well defined map. So this map is well defined. And now we can check that it makes this big sequence exact. And this is mostly fairly straightforward. The only tricky bit is showing that if you've got an element here whose image here is zero, then it's the image of something there. And what you do is just sort of work backwards. First of all, you can lift it to something here, then you can take its image there. Now comes the key point. Since the image of this is zero, you can lift it to something there. Now you can continue all the way back up to there. So anything here whose image is zero is the image of something in here. And the other bits of this diagram are easy to prove. It's easy to prove they're exact. So this diagram is indeed exact and we get the long exact sequence of tour. As an application of the long exact sequence, I just show how to use it. So we might want to use it to find, say tour of Q over Z with Q over Z. And we first note that tour of Q with anything is equal to zero. Here Q is the rationals. I think I forgot to say. And this is because Q is flat. And you remember from the commutative algebra lectures that flatness preserves tensor products. So if you take the resolution naught goes to Z to the M, goes to Z to the N, goes to A, goes to zero and tense with Q or any flat module. The result is exact. And this kernel is tour. So tour vanishes. So now what we can do is we can take the exact sequence naught goes to Z, goes to Q, goes to Q over Z, goes to zero. And we can tensor this with any group A. And we can apply the long exact sequence of tour. So we get tour AQ, goes to tour AQ over Z, goes to Z, tensor A, goes to Z, tensor Q, goes to something. I haven't written out the whole exact sequence. And now this is zero. It's very nice. Now this is just A. And this is just, hang on, it's not Z times A, that's Q tensor A. Q tensor A. And this is tour A with Q over Z. So tour of A with Q over Z is just the kernel of the map from A to Q tensor A. And it's quite easy to work out what this is. The kernel is just the torsion of A. So tour of Q over Z, Q over Z, say, is just the torsion of Q over Z, which is just isomorphic to Q over Z. OK. So far we've been talking about tour for modules over the ring of integers. Next lecture we'll be discussing tour for modules over more complicated rings where things get considerably more complicated.