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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?

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1 Answer 1

up vote 6 down vote accepted

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:

enter image description here

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

enter image description here

is an exact sequence, then you can enlarge it to an exact sequence infinite on both sides:

enter image description here

$ $

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.

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Awesome answer. Very clear, very helpful. Thank ye sir. –  AsinglePANCAKE Oct 5 '12 at 15:54
1  
@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

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