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$\newcommand{\coker}{\operatorname{coker}}\newcommand{\img}{\operatorname{img}}$If $F:M \rightarrow N$ is a map, prove that there is an exact sequence $0 \rightarrow \ker f \rightarrow M \rightarrow N \rightarrow \coker f \rightarrow 0$.

Proof: We need to show that $0 \rightarrow \ker f \rightarrow M$ and $N \rightarrow \coker f \rightarrow 0$ are both exact. Firstly, let $\iota: \ker f \rightarrow M$ be the inclusion. Then $ 0 \rightarrow \ker f \rightarrow M$ is injective since $\ker f $ is exact, i.e. \ker f = \img j = ${0}$ where $j$ sends $0$ to the $\ker f$.Thus exactness follows. Next let $\pi: N \rightarrow \coker f$ be the projection onto the quotient $N/\img (f)$. Then $\coker f $ sends everything in the $\img(f) $ to $0$. So $\img(f) \rightarrow 0$. Thus $N/\img(f) = N/0 \cong N$.

This is where I'm stuck and confused. I'm reading from Rotman's Introduction to Homological Algebra and he doesn't have many problems showing how to deal with these type of problems and in terms of proving, I'm not entirely sure how to approach these problems.

In terms of the above problem I'm confused with whether or not I'm approaching it correctly and if it makes any sense.

Any help or advice would be greatly appreciated.

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  • $\begingroup$ This is correct but you also need to show that $\ker(f) \to M \to N$ and $M \to N \to \text{coker}(f)$ is exact. $\endgroup$ May 18 '18 at 16:30
  • $\begingroup$ I originally thought about doing exactly that but wasn't sure $\endgroup$ May 18 '18 at 16:32
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For what you did, as stated in the comment, restricting the sequence to the two mentioned and proving that they're exact is not enough.

To solve the problem, go back to the very definition of an exact sequence. Do as you said: you consider the inclusion map $\iota:\ker f\to M$, the map $f:M\to N$, the quotient map $\pi:N\to\text{coker} f$ and the zero map $\text{coker}f\to0$. Then, in this order, prove that the image of one is the kernel of the next.

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