Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

1 Answer 1

up vote 0 down vote accepted

Yes, it all looks good, and written in details, this is indeed the straightforward proof that uses the 'generalized element' concept.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.