Question on the exactness at the third kernel $\ker\nu$ in the Snake Lemma

Question

The commutative diagram is as follow:

$\newcommand{\coker}{\operatorname{coker}} \newcommand{\im}{\operatorname{im}} 0\stackrel{}{\longrightarrow} \ker\lambda\stackrel{}{\longrightarrow} \ker\mu\stackrel{}{\longrightarrow} \ker\nu$

$\ \ \ \ \ \ \ \ \ \ \stackrel{}{\downarrow}$ $\ \ \ \ \ \ \ \ \ \ \stackrel{}{\downarrow}$ $\ \ \ \ \ \ \ \ \ \ \stackrel{}{\downarrow}$

$0\stackrel{}{\longrightarrow} L_1\stackrel{\alpha_1}{\longrightarrow} M_1\stackrel{\beta_1}{\longrightarrow} N_1\stackrel{}{\longrightarrow} 0$

$\ \ \ \ \ \ \ \ \ \ \stackrel{\lambda}{\downarrow}$ $\ \ \ \ \ \ \ \ \ \ \stackrel{\mu}{\downarrow}$ $\ \ \ \ \ \ \ \ \ \ \stackrel{\nu}{\downarrow}$

$0\stackrel{}{\longrightarrow} L_0\stackrel{\alpha_0}{\longrightarrow} M_0\stackrel{\beta_0}{\longrightarrow} N_0\stackrel{}{\longrightarrow} 0$

$\ \ \ \ \ \ \ \ \ \ \stackrel{}{\downarrow}$ $\ \ \ \ \ \ \ \ \ \ \stackrel{}{\downarrow}$ $\ \ \ \ \ \ \ \ \ \ \stackrel{}{\downarrow}$

$\coker\lambda\stackrel{}{\longrightarrow} \coker\mu\stackrel{}{\longrightarrow} \coker\nu\stackrel{}{\longrightarrow} 0$

The connection homomorphism is denoted $\delta:\ker\nu\rightarrow\coker\lambda$.

I have trouble varifying the exactness(with arbitrary elements) at $\ker\nu$,i.e. $\beta_1(\ker\mu)=\ker\delta$.

(Notations:$a\in\ker\nu,\delta(a)=f\in\coker\lambda$,mapped through $(a\in\ker\nu)-(b\in N_1)-(c\in M_1)-(d\in M_0)-(e\in L_0)-(f\in\coker\lambda)$.)

Attempt:

It's not hard to show that $\beta_1(\ker\mu)\subset \ker\delta$ :

If we pick $a\in\beta_1(\ker\mu)\subset \ker\nu$,then $\exists c\in M_1,\mu(c)=0\in M_0(c\in \ker\mu)$ since $\beta_1$ is surjective and $\delta$ is well-defined(not relying on how to choose $c$).

Therefore $e=0\in L_0$ since $\alpha_0$ is injective.Then $f=0\in \coker\lambda(\delta(a)=f=0)$,hence $\beta_1(\ker\mu)\subset \ker\delta$.

But I failed dealing with the inverse:

If we trace back along the arrows:$0\in\coker\lambda,\im\lambda\subset L_0,\alpha_0(\lambda(L_1))=\mu\alpha_1(L_1)\subset M_0,\alpha_1(L_1)\subset M_1,\beta_1\alpha_1(L_1)=0\subset N_1$,then $\ker\delta=0$,which makes me confused.

Question:

I think the 'inverse' of $\mu$ from $\mu\alpha_1(L_1)\subset M_0$ is a little bit ambiguous.Can you help me with the proof(of the inverse condition)?

Answer 1

To make things easier to read/write, let us say $\varphi:\ker\mu\rightarrow \ker\nu$ defined as $\varphi=\beta_1|_{\ker\mu}.$ Let us also denote the cosets of $\operatorname{coker}\lambda$ by $[.]_{\lambda}.$

Remember that we can define $\delta$ as $n\mapsto \left[(\alpha_0^{-1}\circ \mu\circ \beta_1^{-1})(n_1)\right]_{\lambda}.$

• Let $n_1\in\ker\delta.$ Then, $\left[(\alpha_0^{-1}\circ \mu\circ \beta_1^{-1})(n_1)\right]_{\lambda}=[0]_{\lambda},$ that is, $(\alpha_0^{-1}\circ \mu\circ \beta_1^{-1})(n_1)\in\operatorname{im} \lambda.$
• So, there is a $l_1\in L_1$ such that $\mu(\beta_1^{-1})(n_1))=\alpha_0(\lambda(l_1))=\lambda(\alpha_1(l_1)),$ that is, $\beta_1^{-1}(n_1)-\alpha_1(l_1)\in\ker \mu.$
• Applying $\beta_1,$ we have that $n_1\in \beta_1(\ker \mu)=\varphi(\ker \mu)$ and so $\ker\delta\subseteq\operatorname{im}\varphi.$