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If $f: M\rightarrow N$ and $g:N\rightarrow P$ such that $g\circ f: M\rightarrow P$ and diagram is commutative

then how we can prove that \begin{equation*} 0\rightarrow \ker(f)\rightarrow \ker(g\circ f)\rightarrow \ker(g)\rightarrow \operatorname{coker}(f)\rightarrow \operatorname{coker}(g\circ f)\rightarrow \operatorname{coker}(g)\rightarrow 0 \end{equation*} is an exact sequence.

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closed as off-topic by Namaste, José Carlos Santos, Arnaldo, Henrik, Xam Sep 27 '17 at 20:11

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  • $\begingroup$ The diagram you have is commutative by definition of $\circ$. $\endgroup$ – Arthur Sep 27 '17 at 6:36
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    $\begingroup$ The obvious diagram to use the snake lemma on, with rows $M\to M\to N$ and $N \to P \to P$ (you want the vertical maps to be $f, g\circ f$ and $g$ respectively) do not have exact rows (at least as long as $M\to M$ and $P\to P$ are identity maps), so you can't use the snake lemma directly on that. $\endgroup$ – Arthur Sep 27 '17 at 6:43
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$\require{AMScd} \newcommand{\coker}{\operatorname{coker}}$ I assume we are in an abelian category (because we are working with kernels, cokernels and exact sequences). Then finite (bi)products exist. I will take the freedom to work with elements. Let $f\colon M\rightarrow N$ and $g\colon N\rightarrow P$ be morphisms; then $g\circ f\colon M\rightarrow P$ is a morphism. Consider the diagram $$ \begin{CD} 0 @>>> M @>a>> M\oplus N @>b>> N @>>> 0\\ & @VfVV @VhVV @VgVV\\ 0 @>>> N @>c>> P\oplus N @>d>> P @>>> 0 \end{CD} $$ where $$ \begin{align} a(m) &= (m,f(m)),\\ b(m,n) &= f(m) -n,\\ c(n) &= (g(n),n),\\ d(p,n) &= p -g(n),\\ h(m,n) &= ((g\circ f)(m), n) \end{align} $$ for all $m\in M$, $n\in N$ and $p\in P$. It is clear that $\ker h \cong \ker (g\circ f)$ and $\coker h \cong \coker (g\circ f)$ canonically. It is also clear that the first and second row are exact. It remains to show that the diagram commutes. For any $m\in M$ and $n\in N$ we compute $$ \begin{align} (h\circ a)(m) &= h(a(m)) = h(m,f(m)) = ((g\circ f)(m),f(m))\\ &= (g(f(m)),f(m)) = c(f(m)) = (c\circ f)(m),\quad\text{and}\\ (g\circ b)(m,n) &= g(b(m,n)) = g(f(m)-n)) = g(f(m)) - g(n)\\ &= (g\circ f)(m) - g(n) = d((g\circ f)(m),n) = d(h(m,n)) = (d\circ h)(m,n). \end{align} $$ Thus, the diagram commutes and we may apply the snake lemma to obtain the desired exact sequence.

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The snake lemma is not immediately applicable as described in my comment above. So instead, here is a check that it is exact at each point.

  1. $\ker f$ is clearly contained in $\ker(g\circ f)$, and inclusion is an injective map.

  2. The map $\ker(g\circ f)\to \ker g$ is given by $f$ restricted to the subset $\ker f\subseteq M$. Clearly an element goes to $0$ here iff it is contained in $\ker f$.

  3. The map $\ker g \to \operatorname{coker} f$ is given by restriction of the canonical projection $N\to N/\operatorname{Im}f$. An element goes to $0$ in the canonical projection iff it is the image of an element in $M$, so we clearly have $\operatorname{Im}\subseteq \ker$ at this point. Reversely, say we have an element $n\in \ker g$ that goes to $0$ in the projection. That means that there is an $m\in M$ with $f(m) = n$. But $g(n) = 0$, so $g(f(m)) = 0$, and thus $m\in \ker(g\circ f)$.

  4. $\operatorname{coker} f \to \operatorname{coker}(g\circ f)$ is the only map that requires real justification for its existence, because it's not simply a restriction or composition. I will name it $\bar g$. Take an element $\bar n\in \operatorname{coker} f$. It has a representative $n_1\in N$. Take $g(n_1)\in P$ and project it down to $\operatorname{coker}(f\circ g)$ to get $\bar g(\bar n)$.
    However, we could've chosen a different representative $n_2$ of $\bar n$. We need to show that this doesn't change which $\bar g(\bar n)$ we end up with. Note that since $n_1$ and $n_2$ both represent $\bar n$, their difference $n_2-n_1$ goes to $0$ in the cokernel. Therefore $n_2-n_1$ is the image of some $m\in M$. That means that $g(n_2) - g(n_1) = g(f(m))$, which goes to $0$ in $\operatorname{coker}(g\circ f)$. So the difference between possible candidates for $\bar g(\bar n)$ is $0$, and they are therefore the same. This makes $\bar g$ well-defined.
    Now for exactness. If an element in $\bar n \in \operatorname{coker}(g\circ f)$ is the image of some $n\in \ker g$, then $n$ is a representative of $\bar n$ as described above. Since $n\in \ker g$, we have $g(n) = 0$, and thus $\bar g(\bar n) = 0$. Reversely, let $\bar n$ be such that $\bar g(\bar n) = 0$. Pick a representative $n\in N$ of $\bar n$. $\bar g(\bar n) = 0$ means that there is some $m\in M$ such that $g(f(m)) = g(n)$. Now note that $n-f(m)$ also represents $\bar n$. However, $g(n-f(m)) = 0$, which means that $n-f(m)\in \ker g$.

  5. We have $\operatorname{Im}(g\circ f)\subseteq \operatorname{Im}g$, which means that $\operatorname{coker}(g\circ f)\to \operatorname{coker}(g)$ is well-defined (it is $P/I \to P/J$ for a $J$ that contains $I$). Take an element $\bar n \operatorname{coker} f$. It has some representative $n\in N$. Clearly $g(n)$ goes to $0$ when we divide out by $\operatorname{Im}g$. Reversely, say we have an element $\bar p \in \operatorname{coker}(g\circ f)$ which goes to $0$ in $\operatorname{coker}g$. That means that a representative $p\in P$ of $\bar P$ is in the image of $g$, so there is some $n\in N$ with $g(n) = p$. Projecting $n$ down into $\operatorname{coker}f$ gives an $\bar n$ which maps to $\bar p$.

  6. The map $\operatorname{coker}(g\circ f)\to \operatorname{coker}(g)$ as described in the paragraph above is clearly surjective, being (isomorphic to) the canonical projection $P/I \to P/I\Big/ J/I$ by the third isomorphism theorem.

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