# Proof of exactness at the first two non-zero objects in the ker-coker sequence (snake lemma).

I am reading MacLane's chapter on Abelian Categories and I am proving the fact, needed for the snake lemma, that the sequence $0\to \text{Ke}f\to \text{Ke}g\to\text{Ke}h$ is exact at $\text{Ke}f$ and $\text{Ke}g$, where $\text{Ke}f$ denotes the domain of the kernel of $f$. MacLane says it follows by an easy diagram chase, but the solution I came up with, involves very simple ideas, yes, but I feel it is a bit lengthy. I just wanted to post what I did to ask if this is the "correct"/expected way to do it.

(I would have added the required diagrams, since it would make life easier to anyone reading, but I do not know how, so please excuse me - Thus I also need to describe all notations).

So, consider two short exact sequences $<m:a\to b,e:b\to c>$, $<m':a'\to b',e':b'\to c'>$ and a morphism $<f:a\to a',g:b\to b',h:c\to c'>$ between them. I also denote the maps between the domains of the kernels by $m_0:\text{Ke}f\to \text{Ke}g$ and $e_0:\text{Ke}g\to \text{Ke}h$.

Firstly, $e_0\circ m_0=0$. Indeed, the upper two squares are commutative, and thus $\text{ker}h\circ e_0\circ m_0=e\circ m\circ\text{ker}f=0$, which implies $e_0\circ m_0=0$, since $\text{ker}h$ is monic.

Next, $m_0$ is monic. Take $x_*\in_m \text{Ke}f$ and suppose that $m_0\circ x_*\equiv 0$. By commutativity, we have $\text{ker}g\circ m_0=m\circ \text{ker}f$, and thus, since $m_0\circ x_*\equiv 0$, we get $m\circ \text{ker}f\circ x_*\equiv 0$. But both $m$ and $\text{ker}f$ are monic, which imply that $x_*\equiv 0$.

Finally, exactness at $\text{Ke}g$. Take $y\in_m \text{Ke}g$, and suppose that $e_0\circ y\equiv 0$. Then, since $\text{ker}h\circ e_0=e\circ \text{ker}g$, we get that $\text{ker}h\circ e_0\circ y=e\circ \text{ker}g\circ y$, and thus $e\circ \text{ker}g\circ y\equiv 0$. Now, by exactness at $b$, there exists $x\in_m a$ such that $m\circ x\equiv \text{ker}g\circ y$. This means that there exist epis $u$ and $v$ such that $\text{ker}g\circ y\circ u=m\circ x\circ v$. Next, we observe that $m'\circ f\circ x\circ v=g\circ m\circ x\circ v=g\circ\text{ker}g\circ y\circ u=0$, and since $v$ is epi we get $m'\circ f\circ x=0$. But $m'$ is monic and so $f\circ x=0$. Thus, $x$ factors through $\text{ker}f$ and therefore there exists $t:\text{dom}(x)\to \text{Ke}f$ such that $x=\text{ker}f\circ t$. Now, $\text{ker}g\circ y\circ u=m\circ x\circ v=m\circ \text{ker}f\circ t\circ v=\text{ker}g\circ m_0\circ t\circ v$, and thus, since $\text{ker}g$ is monic, we get $y\circ u=m_0\circ t\circ v$, or $y\equiv m_0\circ t$, as desired.

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Just as a matter of reference, you can find this fact and lots of others diagram lemmas in § 1.10 of Borcuex' Handbook of Categorical Algebra 2: Categories and Structures . (The snake lemma is fully proved there). – Andrea Gagna Jun 1 '13 at 16:53