# Splicing together Short Exact Sequences

Given an exact sequence of vector spaces $$\cdots\longrightarrow V_{i-1}\longrightarrow V_{i}\longrightarrow V_{i+1}\longrightarrow\cdots$$ I want to show that it is the same as having a collection of short ones such that

$$0\longrightarrow K_i \longrightarrow V_i \longrightarrow K_{i+1}\longrightarrow0$$

So to start I want to show exactness at an arbitrary $V_i$, so I space them suggestively:

$$\begin{array}{c} 0&\rightarrow &K_{i-1}&\rightarrow &V_{i-1}&\rightarrow &K_{i}&\rightarrow&0\\ &&&&0&\rightarrow&K_i&\rightarrow &V_i&\rightarrow&K_{i+1}&\rightarrow&0\\ &&&&&&&&0&\rightarrow&K_{i+1}&\rightarrow&V_{i+1}&\rightarrow&K_{i+2}&\rightarrow&0 \end{array}$$

I drop inclusions down among the corresponding $K_i$'s, and then compose until I get a function from $V_{i-1}$ to $V_i$ and one from $V_i$ to $V_{i+1}$. I check that the image of the first composite mess is the kernel of the second composite mess, which indeed it is.

Question: Am I done? Is showing exactness at one such $V_i$ enough? The question now prompts me to worry about the case were the orginal sequence isn't infinite in both directions...I'm not sure how that case is different?

Well, $V_i$ is an arbitrary object of the putative exact sequence, so if you checked exactness there, that means you're done.

Or, that means you're done checking that the sequence of $V_i$ is an exact sequence, which is not quite what you set out to prove.

You set out to prove that such an exact sequence of $V_i$ "is the same as having a collection of short ones such that...", and yet you merely showed one way: if you have a collection of short exact sequences, you can splice them up.

Now you must show that given a long exact sequence you can chop it up in short exact sequences.

To see that, a diagram is worth a thousand words: where you define $K_n=\operatorname{ker}(f_n)$.

The map $K_n\to A_n$ is the inclusion of the kernel of $f_n$. To get the map $A_{n+1} \to K_n$, observe that:

$K_n=\operatorname{ker}(f_n)=\operatorname{im}(f_{n+1})=\frac{A_{n+1}}{\operatorname{ker}(f_{n+1})}=\frac{A_{n+1}}{\operatorname{im}(f_{n+2})}=\operatorname{coker}(f_{n+2})$.

Hence, define the map $A_{n+1}\to K_n$ as the quotient map of $A_{n+1}$ by the image of $f_{n+2}$.

It is clear by definition that $0\to K_{n+1}\to A_{n+1}\to K_n\to 0$ are short exact sequences.



Now, you worry about how to chop a long exact sequence into short exact sequences. You shouldn't worry, since if is an exact sequence, then you can enlarge it to an exact sequence infinite on both sides: I wrote a couple of pages about splicing and decomposing exact sequences some time ago, which has a couple more results that you might find useful. There is also a section on equivalent definitions of exact functors which exploits these considerations. It is here: it's in Spanish, though.

• Awesome answer. Very clear, very helpful. Thank ye sir. – AsinglePANCAKE Oct 5 '12 at 15:54
• @AsinglePANCAKE: Thanks! you're welcome :) I've added a link at the end that might prove useful. – Bruno Stonek Oct 5 '12 at 17:19
• Cool link! Luckily, I can read spanish pretty well. – AsinglePANCAKE Oct 6 '12 at 17:41