Today I Learned

June 19th

Today I learned a more thorough version of the Snake Lemma to set up for the long exact sequence of cohomology. The following is the main attraction.

The maps on the left and right of the sequence are fairly standard. We cover these quickly. Here are the $\ker$ maps.

We begin by defining our maps.

• The map $\ker\varphi_0\to\ker\varphi_1$ is $\del_M^0$ on restriction. Indeed, any $a\in\ker\varphi_0$ has $\del_M^0\big(\varphi_0(a)\big)=0,$ so $\varphi_1\big(\del_M^0(a)\big)=0$ because of the square. This gives $\del_M^0(a)\in\ker\varphi_1.$

• The map $\ker\varphi_1\to\ker\varphi_1$ is $\del_M^1$ on restriction (again). To repeat, any $a\in\ker\varphi_1$ has $\varphi_1\big(\del_M^0(a)\big)=\del_M^0\big(\varphi_0(a)\big)=0,$ because of the square.

To check the exactness on the left, we need to know $\del_M^0:\ker\varphi_0\to\ker\varphi_1$ surjects onto the kernel of $\del_M^1\ker\varphi_1\to\ker\varphi_2.$ Well, the exactness of the original diagram means\[\ker\del_M^1=\op{im}\del_M^1\]without the restriction. Beginning to introduce the restriction, we note $m_1\in\del_M^0(\ker\varphi_0)$ gives $m_1\in\op{im}\del_M^0=\ker\del_M^1$ for free, so $m_1\in\ker\del_M^1\cap\ker\varphi_1$ from the work above. In the other direction, if $m_1\in\ker\del_M^1\cap\ker\varphi_1=\op{im}\del_M^0\cap\ker\varphi_1,$ we are granted $m_1$ such that\[\del_M^0(m_0)=m_1.\]Using the commuting square, $(\del_N^0\circ\varphi_0)(m_0)=(\varphi_1\circ\del_M^0)(m_0)=\varphi_1(m_1)=0.$ The key step, now, is that $\del_N^1$ is injective in the diagram, so we must have $\varphi_0(m_0)=0,$ implying that, indeed, $m_1\in\del_M^0(\ker\varphi_0).$ Thus, the left side is exact. $\blacksquare$

And here are the $\op{coker}$ maps.

We begin by defining our maps.

• The map $\op{coker}\varphi_0\to\op{coker}\varphi_1$ is induced by $\del_N^0.$ Indeed, $\del_N^0$ takes $N_0\to N_1,$ so we can mod out the image by $\op{im}\varphi_1$ to map into $\op{coker}\varphi_1.$ Now, any $n_0\in\op{im}\varphi_0$ promises $n_0=\varphi_0(m_0)$ so that $(\del_N^0\circ\varphi_0)(m_0)=(\varphi_1\circ\del_M^0)(m_0)\in\op{im}\varphi_1.$ That is, $\del_N^0:\op{im}\varphi_0\to\op{im}\varphi_1,$ meaning we may safely mod the domain by $\op{im}\varphi_0$ to induce $\op{coker}\varphi_0\to\op{coker}\varphi_1.$

• The map $\op{coker}\varphi_0\to\op{coker}\varphi_1$ is induced by $\del_N^0$ in the same way. Again, $\del_N^1\circ\varphi_1=\varphi_2\circ\del_M^1$ implies $\del_N^1:\op{im}\varphi_1\to\op{im}\varphi_2,$ so we safely take $\del_N^1:N_1\to N_2$ and mod out the image by $\op{im}\varphi_2$ and then the domain by $\op{im}\varphi_1.$

We now check the exactness on the right. Again, we start off by knowing that\[\op{im}\del_N^0=\ker\del_N^1,\]so we just have to show that this is maintained after we've done our modding. On one hand, any coset $n_1+\op{im}\varphi_1\in\op{im}\del_N^0$ lands in $\del_N^1(n_1)+\op{im}\varphi_2=0+\op{im}\varphi_2$ after applying $\del_N^1,$ for the bottom row of the diagram is exact.

On the other hand, pick up any coset $n_1+\op{im}\varphi_1\in\ker\del_N^1,$ which we need to show lives in the image $\del_N^0(\op{coker}\varphi_0).$ We have $\del_N^1(n_1)\in\op{im}\varphi_2,$ so we are promised $m_2$ such that $\del_N^1(n_1)=\varphi_2(m_2).$ But $\del_M^1$ is surjective, so we also have $m_1$ such that $\del_N^1(n_1)=(\varphi_2\circ\del_M^1)(m_1).$ But using the square, we see\[\del_N^1(n_1)=\del_N^1\big(\varphi_1(m_1)\big),\]so $n_1-\varphi_1(m_1)\in\ker\del_N^1=\op{im}\del_N^0.$ So we are promised $n_0\in N_0$ such that $\del_N^0(n_0)\in n_1+\op{im}\varphi_1,$ which is what we needed to finish exactness. $\blacksquare$

It remains to talk about $\delta:\ker\varphi_2\to\op{coker}\varphi_0.$ The idea, now, is to pick up an element $m_2\in\ker\varphi_2$ and push it backwards along the diagram to reach $N_0.$ Note we cannot go left twice consecutively in a meaningful way because $\del_\bullet^1\circ\del_\bullet^0=0,$ so we travel $M_2\leftarrow M_1\to N_1\leftarrow N_0.$

Indeed, $\del_M^1$ is surjective, so we are promised an $m_1\in M_1$ such that $\del_M^1(m_1)=m_2.$ This choice is not canonical, but we do have the isomorphism\[\frac{M_1}{\ker\del_M^1}\cong\del_M^1(M_1)=M_2,\]so the choice of $m_1$ is unique up to the (now canonical) coset $m_1+\ker\del_M^1.$ By exactness, we are granted the canonical coset $m_1+\op{im}\del_M^0.$

Continuing, we go through $\varphi_1.$ Note that $\varphi_2(m_2)=0$ means $(\del_N^1\circ\varphi_1)(m_1)=(\varphi_2\circ\del_M^1)(m_1)=0,$ so $\varphi_1(m_1)\in\ker\del_N^1.$ By exactness, we also know $\varphi_1(m_1)\in\op{im}\del_N^0,$ so indeed, we are promised $n_0$ such that $\del_N^0(n_0)=\varphi_1(m_1).$ Note this $n_0$ is unique by the injectivity of $\del_N^0.$

The hope is that, with $m_1$ unique up to choice of coset $m_1+\op{im}\del_M^0,$ we have $n_0$ unique up to $n_0+\op{im}\varphi_0.$ So suppose $m_1'\in m_1+\op{im}\del_M^0$ and pick up $n_0'=(\del_N)^{-1}(\varphi_1(m_1')).$ The point is that\[\del_N^0(n_0-n_0')=\varphi_1(m_1-m_1')\in\op{im}\varphi_1\circ\del_M^0=\op{im}\del_N^0\circ\varphi_0.\]By injectivity, we conclude $n_0-n_0'\in\op{im}\varphi_0$ are indeed unique up to coset of $\op{im}\varphi_0.$ This completes the construction of $\delta.$ To review, our map is\[\left(\del_N^0\right)^{-1}\circ\varphi_1\circ\left(\del_M^1\right)^{-1},\]where we showed the answer is well-defined up to choice of coset $\op{im}\varphi_0$ at the very end.

Now we have to talk about exactness. As a warning, this is about to get messy. Here is the left.

We mostly focus on the kernel of $\delta.$ Elements in the kernel of $\delta$ are those that have made their way into $\op{im}\varphi_0$ by the end, so we manually track $\op{im}\varphi_0$ backwards along\[N_0\stackrel{\del_N^0}\to N_1\stackrel{\varphi_1}\leftarrow M_1\stackrel{\del_M^1}\to M_0.\]We start with $\op{im}\varphi_0\subseteq N_0.$ Moving along $\del_N^0,$ we have $\del_N^0(\op{im}\varphi_0)=\op{im}\del_N^0\circ\varphi_0.$ Using the commutative diagram, this is $\op{im}\varphi_1\circ\del_M^0=\varphi_1(\op{im}\del_M^0),$ and using the exactness of the top row, this is $\varphi_1(\ker\del_M^1).$

Now, we need to move $\varphi_1(\ker\del_M^1)\subseteq M_1$ backwards along $\varphi_1.$ Every single element $\varphi_1(m_1)\in\varphi_1(\ker\del_M^1)$ is going to induce an entire fiber $m_1+\ker\varphi_0$ back in $M_1,$ and unioning over all fibers gives us\[\bigcup_{m_1\in\ker\del_M^1}m_1+\ker\varphi_0=\left\{m_1+m_1':m_1\in\ker\del_M^1\text{ and }m_1'\in\ker\varphi_0\right\}.\]However, when we push this through $\del_M^1,$ every single pair sum $m_1+m_1'$ will have the $m_1\in\ker\del_M^1$ term vanish, leaving us with $\del_M^1(\ker\varphi_1).$

Amazingly, we can check now that the image of $\ker\varphi_1\to\ker\varphi_2,$ induced by $\del_M^1$ by restriction, is exactly $\del_M^1(\ker\varphi_1),$ so the exactness follows. $\blacksquare$

And here is the right.

Again, we begin by focusing on the image of $\delta,$ which we do by tracking forwards along\[M_2\stackrel{\del_M^1}\leftarrow M_1\stackrel{\varphi_1}\to N_1\stackrel{\del_N^0}\leftarrow N_0.\]Because we want the image, we begin with $M_2$ at the left. Because $\del_M^1$ is surjective, the movement along $\del_M^1$ is (boringly) all of $M_1.$ Then we push $M_1$ through $\varphi_1$ to get $\op{im}\varphi_1.$ If we go ahead and push all to $N_0,$ we get $\left(\del_N^0\right)^{-1}(\op{im}\varphi_1).$ We remark that we are actually interested in the cosets $n_0+\op{im}\varphi_0$ for $n_0\in\left(\del_N^0\right)^{-1}(\op{im}\varphi_1).$

Now we talk about the kernel of $\op{coker}\varphi_0\to\op{coker}\varphi_1,$ which is induced by modding out of $\del_N^0.$ Thus, our kernel is the set of cosets $n_0+\op{im}\varphi_0$ such that $\del_N^0(n_0)\in\op{im}\varphi_1.$ This latter condition is equivalent (by the injectivity of $\del_N^0$) to requiring $n_0\in\left(\del_N^0\right)^{-1}(\op{im}\varphi_1),$ so this kernel is the image of $\delta.$ The exactness follows. $\blacksquare$

Putting all of this together, we can splice our exact sequences into a long sequence\[\ker\varphi_0\to\ker\varphi_1\to\ker\varphi_2\stackrel\delta\to\op{coker}\varphi_0\to\op{coker}\varphi_1\to\op{coker}\varphi_2.\]This is exactly what we wanted. $\blacksquare$

With the proof out of the way, I remark that this is not really a proof that should be done in public. There are a very large number of small steps we did above that are true but really annoying to justify, and they're not very instructive anyways.

Regardless, we mention a few related statements, for completeness. Suppose we make the top row of the snake lemma diagram a short exact sequence so that we have the following commutative diagram.

In this case, $\del_M^0$ is injective, so $\ker\varphi_0\to\ker\varphi_1,$ induced by $\del_M^0$ on restriction, is also injective. It follows that we can extend the long exact sequence as\[0\to\ker\varphi_0\to\ker\varphi_1\to\ker\varphi_2\stackrel\delta\to\op{coker}\varphi_0\to\op{coker}\varphi_1\to\op{coker}\varphi_2\]due to this injectivity.

Similarly, suppose we make the bottom of the snake lemma diagram a short exact sequence so that we have the following commutative diagram.

In this case, $\del_N^1$ is surjective, so the map $\op{coker}\varphi_1\to\op{coker}\varphi_2$ (induced by modding $\del_N^1:N_1\to N_2$ by modding) is also surjective. It follows that we can extend the long exact sequence as\[\ker\varphi_0\to\ker\varphi_1\to\ker\varphi_2\stackrel\delta\to\op{coker}\varphi_0\to\op{coker}\varphi_1\to\op{coker}\varphi_2\to0\]due to this surjectivity.

Anyways, while we're here, we mention how to use the Snake lemma to construct the long exact sequence of cohomology.

We begin by saying the map $H^0(G,M)\to H^0(G,N)$ is injective because it is just the map $M^G\to N^G$ induced by $\varphi$ itself on restriction, which is injective by the original short exact sequence. We spend the rest of our time ignoring the $0$ at the start of the long exact sequence.

The strength of the short exact sequence of injective resolutions $0\to I^\bullet\to J^\bullet\to K^\bullet\to0$ is that the following commutative diagram has exact columns; note we have already taken the fixed-point functor.

To extend backwards a bit, we set $\del_\bullet^{-1}$ to be the zero map, and we set $\varphi^{-1}=\varphi$ and $\psi^{-1}=\psi.$

The proof of this statement consists of applying the Snake lemma twice. The first application will give us the long exact sequence, and the second will show that the first is actually valid. Indeed, our first snake diagram is as follows; fix $k\ge0$ a nonnegative integer.

We don't know if the rows are exact yet, so to build suspens for a bit, we will show this in a lemma. Note quickly that the vertical maps are well-defined because $\op{im}\del_\bullet^{k-1}\subseteq\ker\del_\bullet^k$ due to the complex, so these maps are safe.

Note that $(I^k)^G/\op{im}\del_I^{k-1}$ is exactly $\op{coker}\del_I^{k-1},$ and similar for $J$ and $K,$ so we really want the exactness of the sequences\[\begin{cases} \op{coker}\del_I^{k-1}\to\op{coker}\del_J^{k-1}\to\op{coker}\del_K^{k-1}\to0, \\ 0\to\ker\del_I^k\to\ker\del_J^k\to\ker\del_K^k.\end{cases}\]When written this way, this is begging for an application of the Snake lemma. In particular, we have the following diagram, for $\ell\ge0.$

This diagram is just a piece of the short exact sequence of chain complexes, so we exact rows for free. Here the snake lemma (this is the second application of the snake lemma) gives a long exact sequence\[0\to\ker\del_I^\ell\to\ker\del_J^\ell\to\ker\del_K^\ell\to\op{coker}\del_I^\ell\to\op{coker}\del_J^\ell\to\op{coker}\del_K^\ell\to0.\]Setting $\ell=k$ gives the needed $0\to\ker\del_I^k\to\ker\del_J^k\to\ker\del_K^k$ for any $k\ge0.$ Setting $\ell=k-1$ gives the needed $\op{coker}\del_I^{k-1}\to\op{coker}\del_J^{k-1}\to\op{coker}\del_K^{k-1}\to0$ for any $k\ge1.$

It remains to show that the top row is exact for $k=0.$ In this case, $\op{im}\del_\bullet^{k-1}=\op{im}\del_\bullet^{-1}$ is the zero map, so we are really asking for the exactness of $I^0\to J^0\to K^0\to0.$ However, this comes for free from the short exact sequence of complexes, so we are safe. $\blacksquare$

Now that we know our rows are exact, we see the snake lemma gives the induced long exact sequence\[\ker\del_I^k\to\ker\del_J^k\to\ker\del_K^k\to\op{coker}\del_I^k\to\op{coker}\del_J^k\to\op{coker}\del_K^k.\]Here, $\ker\del_\bullet^k$ consists of the cosets $a+\op{im}\del_\bullet^{k-1}$ such that $\del_I^k(a)=0.$ In particular, this kernel is $\ker\del_I^k/\op{im}\del_\bullet^{k-1}=H^k(G,\bullet).$

Further, $\op{coker}\del_\bullet^k$ is $\ker\del_\bullet^{k+1}$ modded by the image of $\del_\bullet^k$ under $(\bullet^k)^G.$ (The modding by $\op{im}\del_\bullet$ doesn't matter because all these elements go to $0.$) Thus, the cokernel is $\ker\del_\bullet^{k+1}/\op{im}\del_\bullet^k=H^{k+1}(G,\bullet).$ This cokernel computation is why our bottom row consits of kernels.

Bringing this all together, we have the long exact sequence\[H^k(G,M)\to H^k(G,N)\to H^k(G,P)\to H^{k+1}(G,M)\to H^{k+1}(G,N)\to H^{k+1}(G,N).\]We also note the first two maps $H^k(G,N)\to H^k(G,N)\to H^k(G,P)$ are induced by $\varphi^k$ and $\psi^k$ by restriction, so they are really the map $\varphi^k:(I^k)^G\to(J^k)^G$ and $\psi^k:(J^k)^G\to (K^k)^G$ upon modding out in the codomain and domain. This is to say that the maps on the left are the good ones $H^k(\varphi)$ and $H^k(\psi).$

Similarly, the last two maps $H^{k+1}(G,N)\to H^{k+1}(G,N)\to H^{k+1}(G,N)$ are induced by the restrictions of $\varphi^{k+1}:\ker\del_I^{k+1}\to\ker\del_J^{k+1}$ and $\psi^{k+1}:\ker\del_J^{k+1}\to\ker\del_K^{k+1}$ after modding out the codomain and domain by some images. Again, this tells us the maps on the right are the good ones $H^{k+1}(\varphi)$ and $H^{k+1}(\psi)$ by construction.

Thus, we have a series of long exact sequences for $k\ge0,$ as follows; here, $\delta^k$ is given by the Snake lemma.

Because of each these sequences are exact on their own, and the sequence for $k$ and $k+1$ share the bottom and top row respectively (that is, the maps are the same), we may splice all of these sequences together into the desired long exact sequence. $\blacksquare$

As some commentary, it might feel very strong that we required the short exact sequence of complexes, and indeed, this is not easy to prove, and I do not yet know its proof. However, it is always possible. We say (but do not prove) that merely fixing the injective resolution $J^\bullet$ of $N$ to be $I^\bullet\oplus K^\bullet$ will do the trick.