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I encountered the following exact sequence a while ago, and wondered if it was a special case of the Snake lemma. It looks like it would be, but I don't quite see how...

The context is that $A,B$ are operators on a Hilbert space $H$ (but it holds in greater generality):

$$\DeclareMathOperator{\Im}{Im} 0 \rightarrow \ker(B) \rightarrow \ker(AB) \xrightarrow{B} \ker(A) \rightarrow H/\Im(B) \xrightarrow{A} H/\Im(AB) \rightarrow H/\Im(A) \rightarrow 0$$

alexwlchan
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Tony
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3 Answers3

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\begin{matrix} 0\to &H &\stackrel{(1,B)}{\longrightarrow} &H\oplus H &\stackrel{B\pi_1-\pi_2}{\longrightarrow} &H &\to0\\ &\downarrow\rlap{\small B} & &\downarrow\rlap{\small(AB,1)} & &\downarrow\rlap{\small A} \\ 0\to&H &\stackrel{(A,1)}{\longrightarrow} &H\oplus H &\stackrel{\pi_1-A\pi_2}{\longrightarrow} &H&\to0 \end{matrix} This should work.

Sungjin Kim
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In general, let $\alpha: L \rightarrow M$ and $\beta: M \rightarrow N$ be morphisms in an abelian category. Then

$\begin{matrix} &L &\stackrel{\alpha}{\longrightarrow} &M &\stackrel{}{\longrightarrow} &\operatorname{coker} \alpha &\to0\\ &\downarrow\rlap{\beta \alpha} & &\downarrow\rlap{\beta} & &\downarrow\rlap{} \\ 0\to&N &\stackrel{\text{id}}{\longrightarrow} &N &\stackrel{}{\longrightarrow} &0 \end{matrix}$

yields an exact sequence $$\ker \beta \alpha \rightarrow \ker \beta \rightarrow \operatorname{coker} \alpha \rightarrow \operatorname{coker} \beta\alpha \rightarrow \operatorname{coker} \beta \rightarrow 0$$

and

$\begin{matrix} &0 &\stackrel{}{\longrightarrow} &L &\stackrel{\text{id}}{\longrightarrow} &L &\to0\\ &\downarrow\rlap{} & &\downarrow\rlap{\alpha} & &\downarrow\rlap{\beta \alpha} \\ 0\to& \ker \beta &\stackrel{}{\longrightarrow} &M &\stackrel{\beta}{\longrightarrow} &N \end{matrix}$

yields an exact sequence

$$0 \rightarrow \ker \alpha \rightarrow \ker \beta \alpha \rightarrow \ker \beta \rightarrow \operatorname{coker} \alpha \rightarrow \operatorname{coker} \beta\alpha \, .$$

It is not hard to check that the two sequences can be combined to give $$0 \rightarrow \ker \alpha \rightarrow \ker \beta \alpha \rightarrow \ker \beta \rightarrow \operatorname{coker} \alpha \rightarrow \operatorname{coker} \beta\alpha \rightarrow \operatorname{coker} \beta \rightarrow 0 \, .$$

Cihan
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  • You don't need to apply the snake lemma twice. People rarely mention this, but it is true in general that the kernel of the first morphism in the ker-coker sequence is isomorphic to the kernel of the first morphism in the first exact sequence, so you can get the start $0 \to \ker{\alpha} \to \ker{\beta\alpha}$ from that. – Martin Jul 06 '13 at 09:26
  • @Martin: Sure, and dually, it is also true in general that $\operatorname{coker} \beta$ is a cokernel of $\operatorname{coker} \alpha \rightarrow \operatorname{coker} \beta \alpha$. I just wanted to mention that there are two diagrams for which the snake lemma gives at once almost all of the desired long exact sequence. – Cihan Jul 06 '13 at 10:53
  • I see. I guess I didn't make myself very clear in my first comment. What I was trying to say is this: In the snake sequence of $$ \begin{matrix} &A &\stackrel{f}{\longrightarrow} &B &\stackrel{g}{\longrightarrow} &C &\to0\ &\downarrow\rlap{a} & &\downarrow\rlap{b} & &\downarrow\rlap{c} \ 0\to&A' &\stackrel{f'}{\longrightarrow} &B' &\stackrel{g'}{\longrightarrow} &C' \end{matrix} $$ one has that $\ker{f}$ is isomorphic to the kernel of the map $\ker{a} \to \ker{b}$ (and dually $\operatorname{coker}{g'}$ is isomorphic to the cokernel of $\operatorname{coker}b \to \operatorname{coker}c$). – Martin Jul 06 '13 at 11:34
  • Oh now I see. In fact I didn't realize this situation in the general setup of the snake lemma before. Thanks! – Cihan Jul 06 '13 at 11:44
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I think this should work: $$ \begin{matrix} &H &\stackrel{(-B,1)}{\longrightarrow} &H\oplus H &\stackrel{\pi_1+B\pi_2}{\longrightarrow} &H &\to0\\ &\downarrow\rlap{\small B} & &\downarrow\rlap{\small(AB,-AB)} & &\downarrow\rlap{\small A} \\ 0\to&H &\stackrel{(-AB,-A)}{\longrightarrow} &H\oplus H &\stackrel{}{\longrightarrow} &H \end{matrix} $$

Oops, after writing all down, it doesn't look that good anymore ... May have to sleep over it

Hagen von Eitzen
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  • What is the map on the bottom right? – Sungjin Kim Jul 04 '13 at 16:23
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    Bottom left map should be injective, but it is not in yours. – Sungjin Kim Jul 04 '13 at 16:54
  • Bottom right could be $\pi_1-B\pi_2$. But then the right square is no more commutative. And indeed the bottom left is not injective without certain conditions. But there are verisimilarly some modifications which could solve the problem. – awllower Jul 05 '13 at 05:07