Today I Learned

June 17th

Today I learned that group cohomology is well-defined. Recall our definition from a two days ago.

Yesterday we completed the proof that every $G$-module has an injective resolution, which completed the proof of the existence of the cohomology. Today we show that $H(G,M)$ does not depend on our choice of injective resolution.

We will want the homorphism theorem for $G$-modules, so we quickly get it out of the way. This holds in arbitrary abelian categories, but I don't know the definition of an abelian category, so we will just show this.

We start by showing that $\ker\varphi$ is a $G$-submodule of $M.$ Looking at the underlying abelian groups, it is already a subgroup. To show that it is fixed by the action of $G,$ pick up any $g\in G$ and $m\in\ker\varphi$ and observe\[\varphi(gm)=g\cdot\varphi(m)=g\cdot 0_N=0_N,\]so indeed, $gm\in\ker\varphi.$ (I guess $g\cdot0=0$ because $g\cdot0=g\cdot(0+0)=g\cdot0+g\cdot0.$) It follows that $M/\ker\varphi$ is defined as a $G$-module: it is already an abelian group, and its group action is\[g\left(a+\ker\varphi\right)=ga+\ker\varphi,\]which is well-defined because $a+\ker\varphi=b+\ker\varphi,$ which implies $a-b\in\ker\varphi,$ which implies $a-b\in\ker\varphi,$ which implies $ga-gb\in\ker\varphi,$ which implies $ga$ and $gb$ are in the same coset.

To get the isomorphism asserted, we send $m+\ker\varphi\in M/\ker\varphi$ to $\varphi(m).$ By group theory, this is already a group isomorphism, so we only have to check that the group action is preserved. Picking up a $g\in G,$ we note\[g(m+\ker\varphi)=gm+\ker\varphi\mapsto\varphi(gm)=g\cdot\varphi(m),\]so indeed, this is a $G$-module homomorphism. Finally, we note that $\varphi(M)$ is immediately a $G$-module because it is isomorphic to one. This finishes the proof. $\blacksquare$

Our main tool showing that $H^\bullet(G,M)$ is well-defined will be the fact that a $G$-module homomorphism $f:M\to N$ induces a map between cohomology groups.

Note that to prove the diagram commutes, we merely have to prove that each square commutes. We can show this inductively, but is is more directly true that two consecutive arrows in the top or bottom complex give the $0$ mapping, so they will commute just fine. We will not dwell on this.

Anyways, we do induction on $k.$ By convention, we will set $M=I^{-1}$ and $N=K^{-1}$ so that $\varphi^{-1}:=\varphi.$ Additionally, we set $\del_I^{-2}:0\to M$ and $\del_J^{-2}:0\to N$ with $\del_I^{-1}:M\to I^0$ and $\del_J^{-1}:N\to J^0$ to be the embeddings. (Our $\del_\bullet^{-1}$ are not the zero mapping here!) Observe that we can also realize these as first modding out $I^{-1}$ by "$I^{-2}=0$'' and then embedding into an injective $G$-module, just like the higher $\del^k$s.

For the inductive step, we are allowed to assume the previous square commutes (for $k=-1,$ this square includes $0\to 0$) and that $I^{k+1}$ and $J^{k+1}$ are injective $G$-modules. (For $k=-1,$ $I^k$ and $J^k$ are not injective.) We need to induce $\varphi^{k+1}:I^{k+1}\to J^{k+1}.$ For this, we decompose $\del^k_I$ and $\del^k_J$ into modding followed by embedding.

As for why we can do this, we note that $\del^k$ has kernel exactly $\del^{k-1}\left(I^{k-1}\right)$ because $I^{k-1}\to I^k\to I^{k+1}$ is exact, so we are allowed to mod out by this image, and after we do, we have an isomorphism\[\del^k_I\left(I^k\right)\cong\frac{I^k}{\ker\del^k_I}=\frac{I^k}{\del^{k-1}_I\left(I^{k-1}\right)}\]induced by the $G$-module homomorphism $\del^k.$ The same holds for $J.$ This means that after the modding out, we have an isomorphism into the image, which is the same thing as an embedding.

Anyways, we now induce $\psi^k$ and $\varphi^{k+1}.$ We can induce $\psi^k$ directly from the previous square.

(Our base case is still valid because $\varphi^{k-1}:I^{k-1}\to J^{k-1}$ is $0\to 0.$) We begin creating $\psi^k$ from the composite\[I^k\stackrel{\varphi^k}\to J^k\to J^k/\del_J^{k-1}\left(J^{k-1}\right).\]Now, we want to show that this mapping is defined up to coset of $\del^{k-1}_I\left(I^{k-1}\right),$ so we have to show that the image $\del_I^{k-1}\left(I^{k-1}\right)$ lives in the kernel of the above map. Well, the commutativty of the previous square says\[\varphi^k\left(\del_I^{k-1}\left(I^{k-1}\right)\right)=\del_J^{k-1}\left(\varphi^{k-1}\left(I^{k-1}\right)\right)\subseteq\del_J^{k-1}\left(J^{k-1}\right),\]so indeed, we may mod out by $\del_I^{k-1}\left(I^{k-1}\right).$ That is, $\psi^k$ is just $\varphi^k$ after modding out the domain and codomain.

We take a moment to verify that the constructed square commutes.

Indeed, $\psi^k$ is defined to take any coset $a+\del_I^{k-1}\left(I^{k-1}\right)$ directly to $\varphi(a)+\del_J^{k-1}\left(J^{k-1}\right),$ where we just spend the previous paragraph showing well-definedness. So going around the top or bottom of the square sends $a\in I^k$ to $\varphi(a)+\del_J^{k-1}\left(J^{k-1}\right).$

It remains to induce $\varphi^{k+1}.$ For this we focus on the right half of our original diagram, as follows.

We now induce $\varphi^{k+1}$ from the injectivity of $I^{k+1}$ and $J^{k+1}.$ Indeed, the top embedding lets us write (abusing notation) $I^k/\del_I^{k-1}\left(I^{k-1}\right)\subseteq I^{k+1},$ and we already have a $G$-module homomorphism $I^k/\del_I^{k-1}\left(I^{k-1}\right)\to J^k/\del_J^{k-1}\left(J^{k-1}\right)\to J^{k+1}.$ So we can extend this $G$-module homomorphism to all of $I^{k+1}$ by the injectivity of $J^{k+1},$ which is our $\varphi^{k+1}.$

We now finish by showing that the square we just constructed commutes. Then the inductive step will be done by adjoining the two commuting squares, which will create the desired commuting square. Namely, we show the following commutes.

Indeed, $\varphi^{k+1}$ is defined as extending the path through the right of the square. Explicitly, $a\in I^k/\del_I^{k-1}\left(I^{k-1}\right)\subseteq I^{k+1}$ is sent to the same place along $I^k/\del_I^{k-1}\left(I^{k-1}\right)\to J^k/\del_J^{k-1}\left(J^{k-1}\right)\to J^{k+1}$ as along $\varphi^{k+1}.$ This finishes the proof. $\blacksquare$

The above construction is so important, it has a name.

These induced maps in fact generate a map between cohomology groups, as promised.

As discussed a few days ago, a $G$-module homomorphism induces a map between the $G$-fixed points of the domain and codomain. So we have the following commutative diagram created by restricting the hypothesized chain morphism.

Now, we need to induce a map $H^k(G,M)\to H^k(G,N),$ which given these injective resolutions, means we need a map from $\ker\del_I^k/\op{im}\del_I^{k-1}$ to $\ker\del_J^k/\op{im}\del_J^{k-1}.$

To begin, we consider the following commuting square.

Note that $\ker\del_I^k$ vanishes along $\varphi^{k+1}\circ\del_I^k,$ so it must also vanish along $\del_J^k\circ\varphi^k.$ This means that $\varphi^k$ will restrict to a map $\ker\del_I^k\to\ker\del_J^k,$ so we hope that our $\varphi^k$ will induce a

Next, we look at the other side.

Now, we notice that the image of $\varphi^k\circ\del_I^{k-1}$ is equal to the image of $\del_J^{k-1}\circ\varphi^{k-1}$ and hence a subset of the image $\del_J^{k-1}.$ Thus, $\varphi^k$ takes the image of $\del_I^{k-1}$ to the image of $\del_J^{k-1}$ on restriction.

Before continuing, we take a second to codify what just happened.

This follows from the above discussion. $\blacksquare$

Anyways, bringing this together, we take the restricted map\[\varphi^k:\ker\del_I^k\to\ker\del_J^k\]and mod out the image by the $G$-(sub)module $\op{im}\del_J^{k-1}.$ Because now $\varphi^k$ (under restriction) takes $\op{im}\del_I^{k-1}$ into the kernel of $\op{im}\del_J^{k-1},$ we are guaranteed a well-defined map\[\ker\del_I^k/\op{im}\del_I^{k-1}\to\ker\del_J^k/\op{im}\del_J^{k-1}\]induced by $\varphi^k.$ This is exactly what we wanted. $\blacksquare$

What's amazing about the induced map of cohomology groups is that they are unique up to choice of $\varphi.$ Namely, exactly which $\varphi^\bullet$ we chose is irrelevant.

Before doing anything, we start by saying when $\bullet=0,$ the map of cohomology groups $H^0(G,M)=M^G$ to $H^0(G,N)=N^G$ induced by either $\varphi_1^\bullet$ or $\varphi_2^\bullet$ is the same because this is just the restriction of $\varphi.$ We say this now because we are about to set an unusual convention to $\del_\bullet^{-1}$ (namely, we make it the embedding, not the zero map) in order to make our diagram commute nicer.

Our first goal will be to decrease the number of arrows. Extend all of $I^k,J^k,\varphi^k_\bullet,\del_\bullet^k$ backwards to $k=-1$ and $k=-2,$ where $\varphi_1^{-1}=\varphi_2^{-1}=\varphi$ and $\varphi_1^{-2}=\varphi_2^{-2}:0\to 0.$ The commutative diagram tells us that $\del_J^k\circ\varphi_\bullet^k=\varphi_\bullet^{k+1}\circ\del_I^k$ for any $k\ge-2,$ which implies that, upon subtracting,\[\del_J^k\circ\left(\varphi_2^k-\varphi_1^k\right)=\left(\varphi_2^{k+1}-\varphi_1^{k+1}\right)\circ\del_I^k.\]Indeed, the sum/difference of two $G$-module homomorphisms $\varphi_1:A\to B$ and $\varphi_2:A\to B$ is another $G$-module homomorphism because the underlying group homomorphisms work, and for any $g\in G$ and $a\in A,$ $(\varphi_1-\varphi_2)(ga)=\varphi_1(ga)-\varphi_2(ga)=g\cdot(\varphi_1(a)-\varphi_2(a))$ by linearity. Then the fact we have a group homomorphism means, for any $a\in I^k,$\[\begin{cases} \left(\del_J^k\circ\left(\varphi_2^k-\varphi_1^k\right)\right)(a)=\left(\del_J^k\circ\varphi_2^k\right)(a)-\left(\del_J^k\circ\varphi_1^k\right)(a), \\ \left(\left(\varphi_2^{k+1}-\varphi_1^{k+1}\right)\circ\del_I^k\right)(a)=\left(\varphi_2^{k+1}\circ\del_I^k\right)(a)-\left(\varphi_1^{k+1}\circ\del_I^k\right)(a),\end{cases}\]so the commutativity of the diagram gives the assertion. We remark that the above was merely showing that morphisms form a ring of composition over addition, which is checked when we say the category of $G$-modules is abelian. We will not show this explicitly again.

Recollecting our morphisms, the following diagram commutes.

This diagram does indeed have fewer arrows, so we would like to work with it instead. Recall that the maps $\varphi_\bullet^k$ induces of $H^k(G,M)\to H^k(G,N)$ were simply $\varphi_\bullet^k$ restricted and then modded. So if we want to show $\varphi_1^k$ and $\varphi_2^k$ induce the same map, it suffices to show that $\varphi_2^k-\varphi_1^k$ induces (i.e., restricts to) the zero map.

Thus, it suffices to show the following. Note $\varphi-\varphi=0.$

The idea is to create a chain homotopy. Again, extend all of $I^k,J^k,\varphi^k_\bullet,\del_\bullet^k$ backwards to $k=-1$ and $k=-2.$ We claim that we exhibit maps $\psi^k:I^{k+1}\to J^k$ for $k\ge-2$ such that $\varphi^k=\psi^k\circ\del_I^k+\del_J^{k-1}\circ\psi^{k-1}$ for $k\ge-1.$ Here is the diagram.

As a warning, the dashed arrows do not commute with the other arrows.

We construct the $\psi^k$ to satisfy $\varphi^k=\psi^k\circ\del_I^k+\del_J^{k-1}\circ\psi^{k-1}$ by induction. We begin with $\psi^{-2}=0$ for free (it is the zero map), and our base case is $\psi^{-1}=0,$ which works because $0=0\circ\del_I^k+\del_J^{k-1}\circ0$ is a true statement. For the inductive step, we suppose we have $\varphi^k=\psi^k\circ\del_I^k+\del_J^{k-1}\circ\psi^{k-1},$ and we construct $\psi^{k+1}.$

For this, we use the long exact sequence to decompose $\del_I^{k+1}:I^{k+1}\to I^{k+2}$ into $I^{k+1}\to I^{k+1}/\op{im}\del_I^k\hookrightarrow I^{k+2}$ in the same lemma establishing the lifting. (Explicitly, $\del_I^{k+1}$ induces the isomorphism $I^{k+1}/\op{im}\del_I^k=I^{k+1}/\ker\del_I^{k+1}\cong\op{im}\del_I^{k+1}.$) Here are the morphisms we care about.

The hard part will be done attempting to induce $s:I^{k+1}/\op{im}\del_I^k$ such that\[\varphi^{k+1}\stackrel?=s\circ d+\del_J^k\circ\psi^k.\]Indeed, if we can induce such an $s:I^{k+1}/\op{im}\del_I^k,$ then we note $I^{k+1}/\op{im}\del_I^k\subseteq I^{k+2}$ means injecivity gives us the desired map $\psi^{k+1}:I^{k+2}\to J^{k+1}$ which restricts to $s$ on $I^{k+1}/\op{im}\del_I^k.$ Namely, we will still have\[\varphi^{k+1}=\psi^{k+1}\circ\del_I^{k+1}+\del_J^k\circ\psi^k\]because the image of $\del_I^{k+1}$ is the image of $d,$ on which $\psi^k$ restricts to $s.$

So we now attempt to construct $s.$ Rearranging, we see we need\[s\circ d\stackrel?=\varphi^{k+1}-\del_J^k\circ\psi^k.\]Because $d$ merely mods out by $\op{im}\del_I^k,$ it suffices to show that $\varphi^{k+1}-\del_J^k\circ\psi^k$ as a map $I^{k+1}\to J^{k+1}$ has $\op{im}\del_I^k$ in its kernel so that we can mod out the domain $I^{k+1}$ by $\op{im}\del_I^k$ safely. That is, we need\[\left(\varphi^{k+1}-\del_J^k\circ\psi^k\right)\circ\del_I^k\]is the $0$ mapping. Using the fact our moprhisms form a ring of composition over addition, we see that we need to show\[\varphi^{k+1}\circ\del_I^k\stackrel?=\del_J^k\circ\psi^k\circ\del_I^k.\]To finish, we expand out $\psi^k.$ Recall $\varphi^k=\psi^k\circ\del_I^k+\del_J^{k-1}\circ\psi^{k-1},$ so $\psi^k\circ\del_I^k=\varphi^k-\del_J^{k-1}\circ\psi^{k-1}.$ Substituting, we see\[\del_J^k\circ\psi^k\circ\del_I^k=\del_J^k\left(\varphi^k-\del_J^{k-1}\circ\psi^{k-1}\right)=\del_J^k\circ\varphi^k-\del_J^k\circ\del_J^{k-1}\circ\psi^{k-1}.\]Note $\del_J^k\circ\varphi^k=\varphi^{k+1}\circ\del_I^k$ by construction of the $\varphi^\bullet$ as a chain morphism, and this is our desired left-hand side. As for $\del_J^k\circ\del_J^{k-1}\circ\psi^{k-1},$ we see $\del_J^k\circ\del_J^{k-1}$ is the $0$ mapping by the exactness of our injective resolution, so it vanishes. This completes the justification of the existence of $s,$ and so the inductive step, and so the construction of the $\psi^\bullet.$

It remains to use the chain homotopy to show that the induced map on the cohomology groups is the zero mapping. We go ahead and apply the fixed-point functor everywhere, but all of our maps are preserved because they were $G$-module homomorphisms (and are now group homomorphisms). Recall that $H^k\left(\varphi^\bullet\right):H^k(G,M)\to H^k(G,N)$ was created by noting $\varphi^k$ restricts to a map\[\varphi^k:\ker\del_I^k\to\ker\del_J^k\]and then modding out the image by $\op{im}\del_J^{k-1}$ (and modding out the domain by $\op{im}\del_I^{k-1}$). Explicitly, we have that $H^k\left(\varphi^\bullet\right)\left(a+\op{im}\del_I^{k-1}\right):=\varphi(a)+\op{im}\del_J^{k-1}.$

So to show that our $H^k\left(\varphi^\bullet\right)$ is the $0$ map, it suffices to show that this restriction of $\varphi^k$ maps into $\op{im}\del_J^{k-1}.$ Well, pick up any $a\in\ker\del_I^k,$ and the chain homotopy implies\[\varphi^k(a)=\left(\psi^k\circ\del_I^k\right)(a)+\left(\del_J^{k-1}\circ\psi^{k-1}\right)(a).\]Because $a\in\ker\del_I^k,$ the first summand evaporates, and the second summand is $\del_J^{k-1}\left(\psi^{k-1}(a)\right)$ is in the image of $\del_J^{k-1}.$ So indeed, $\varphi^k$ does indeed take $\ker\del_I^k\to\op{im}\del_J^{k-1}.$ All of these maps are well-defined and work as discussed after applying the fixed-point functor as well. $\blacksquare$

As discussed preceding the above lemma, the lemma was enough to complete the proof. $\blacksquare$

In light of the previous lemma, we begin to notate the induced $H^\bullet\left(\varphi^\bullet\right):H(G,M)\to H(G,N)$ from the $G$-module homomorphism $\varphi:M\to N$ by just $H^\bullet(\varphi)$ because the exact choice of lift $\varphi^\bullet$ doesn't matter, established above.

Before continuing, we remark that we are beginning to get the feeling that $H^\bullet$ might actually be a functor from the category of $G$-modules to abelian groups. Indeed, it sends a $G$-module $M$ to $H^\bullet(G,M)$ and a morphism $\varphi$ to $H^\bullet(\varphi).$ To finish establishing this, we need to show the following.

For this, we have to go back to the definition of $H^\bullet(\varphi)$ and $H^\bullet(\psi).$ Fix lifts of $\varphi$ to $\varphi^\bullet:I^\bullet\to J^\bullet$ and $\psi^\bullet:J^\bullet\to K^\bullet.$ For concreteness, here is our diagram.

We recall that $H^\bullet(\varphi)$ was defined by applying the fixed-point functor and then restricting $\varphi^k$ to $\ker\del_I^k\to\ker\del_J^k$ followed by modding out the codomain by $\op{im}\del_J^{k-1}$ and then the domain by $\op{im}\del_I^{k-1}.$ Of course, $\psi^k$ has a similar story.

Thus, we note that we have the defining equations (here $\del^{-1}_\bullet$ is the zero map!)\[\begin{cases} H^\bullet(\varphi)\left(a+\op{im}\del_I^k\right):=\varphi(a)+\op{im}\del_J^k, \\ H^\bullet(\psi)\left(b+\op{im}\del_J^k\right):=\psi(b)+\op{im}\del_K^k, \\ H^\bullet(\psi\circ\varphi)\left(a+\op{im}\del_I^k\right):=(\varphi\circ\psi)(b)+\op{im}\del_K^k.\end{cases}\]From these definitions we see that, indeed, $H^\bullet(\psi)\circ H^\bullet(\varphi)$ and $H^\bullet(\psi\circ\varphi)$ do the same thing to $a+\op{im}\del_I^k,$ so we are done here. $\blacksquare$

We are now ready to show that our cohomology groups are well-defined. The intuition at this point is that our induced map of cohomology groups by a morphism $\varphi$ is "canonical,'' and this we expect to be enough to show that this map is an isomorphism in special cases. Here is the conversion.

This is reasonably slick. Set up the chain morphisms on top of each other.

By construction of the $\varphi^\bullet$ and $\psi^\bullet,$ all of the square commute, so all of the $2\times1$ rectangles. What we see, them, is that $\psi^\bullet\circ\varphi^\bullet$ lifts $\psi\circ\varphi=\op{id}_M:M\to M$ as a chain morphism.

The kicker is that $\op{id}^\bullet:I^k\to I^k$ also lifts $\op{id}_M:M\to M,$ but we know that the induced map $H^\bullet(\op{id}_M)$ takes the coset $a+\op{im}\del_I^\bullet$ directly to $a+\op{im}\del_I^\bullet,$ which is $\op{id}_{H^\bullet(G,M)}.$ Because $\psi^\bullet\circ\varphi^\bullet$ and $\op{id}^\bullet$ both lift $\op{id},$ we conclude\[H^\bullet(\psi)\circ H^\bullet(\varphi)=H^\bullet(\psi\circ\varphi)=H^\bullet(\op{id})=\op{id}_{H^\bullet(G,M)}\]using that funtoriality of $H^\bullet$ we just so happened to need. This finishes. $\blacksquare$

And here is the main attraction.

Note that $M\cong N$ promises us maps $\varphi:M\to N$ and $\psi:N\to M$ such that $\psi\circ\varphi=\op{id}_M$ and $\varphi\circ\psi=\op{id}_N.$ Then the previous proposition implies\[H^\bullet(\psi)\circ H^\bullet(\varphi)=\op{id}_{H^\bullet(G,M)},\quad\text{and}\quad H^\bullet(\varphi)\circ H^\bullet(\psi)=\op{id}_{H^\bullet(G,N)}.\]Thus, $H^\bullet(\varphi)$ and $H^\bullet(\psi)$ witness $H^\bullet(G,M)\cong H^\bullet(G,N),$ and we are done. $\blacksquare$

In particular, if $M=N,$ then $\op{id}_M$ witnesses our isomorphism $M\cong M,$ so it doesn't matter what choice of injective resolutions we have, our group cohomology is always the same. This is perhaps remarkable because we can reliably make an injective module bigger, so the injective resolution is potentially very large. However, the moderating factor is that the definition of cohomology is modding in a long exact sequence, and these mods accommodate for steep size increases.