Edit: I got stuck in a couple of places in proving the statement below. Specifically:
- $d$ is well-defined.
- $\text{ker}(d) \subseteq \text{im}(\bar{v})$
- $\text{ker}(\bar{v}’) = \text{im}(\bar{u}’)$
My attempt so far:
- well-defineness of $d$: as below
- $\text{ker}(d) \subseteq \text{im}(\bar{v})$: let $m’’ \in \text{ker}(f)$. Since $v$ is surjective, there exists $m \in M$ such that $v(m) = m’’$. Now I need to show that $m \in \text{ker}(f)$. I also got $n’ \in N’$ s.t. $u’(n’) = f(m)$, and $m’ \in M’$ s.t. $f’(m’) = n’$, but can’t go further.
- In proving $\text{ker}(\bar{v}’) \subseteq \text{im}(\bar{u}’)$, for $[n] \in \text{ker}(\bar{v}’)$, I’m not sure what I can say about the statement $\bar{v}’([n]) = 0_{\text{coker}(f’’)} = [\text{im}(f’’)]$.
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I need to check that the map $d$ in the following is well-defined.
Proposition 2.10. Let
be a commutative diagram of $A$-modules and homomorphisms, with the rows exact. Then there exists an exact sequence $$ \begin{aligned} 0 \rightarrow \operatorname{Ker}\left(f^{\prime}\right) \stackrel{\bar{u}}{\rightarrow} \operatorname{Ker}(f) \stackrel{\bar{v}}{\rightarrow} \operatorname{Ker}\left(f^{\prime \prime}\right) \stackrel{d}{\rightarrow} & \text { Coker }\left(f^{\prime}\right) \stackrel{\bar{u}’}{\rightarrow} \operatorname{Coker}(f) \stackrel{\bar{v}’}{\rightarrow} \operatorname{Coker}\left(f^{\prime \prime}\right) \rightarrow 0 \end{aligned} $$ in which $\bar{u}, \bar{v}$ are restrictions of $u, v$, and $\bar{u}^{\prime}, \bar{v}^{\prime}$ are induced by $u^{\prime}, v^{\prime}$.
The boundary homomorphism $d$ is defined as follows: if $x^{\prime \prime} \in \operatorname{Ker}\left(f^{\prime \prime}\right)$, we have $x^{\prime \prime}=v(x)$ for some $x \in M$, and $v^{\prime}(f(x))=f^{\prime \prime}(v(x))=0$, hence $f(x) \in \operatorname{Ker}\left(v^{\prime}\right)=$ $\operatorname{Im}\left(u^{\prime}\right)$, so that $f(x)=u^{\prime}\left(y^{\prime}\right)$ for some $y^{\prime} \in N^{\prime}$. Then $d\left(x^{\prime \prime}\right)$ is defined to be the image of $y^{\prime}$ in Coker $\left(f^{\prime}\right)$.
I think the issue is that there might exist $\chi \in M, \chi \neq x$ such that $v(\chi) = v(x) = x’’$. I need to show that the value $d(x’’)$ obtained via $\chi$ is the same as the one obtained via $x$.
So assume that such $\chi$ exists. We’d have $v’(f(\chi)) = f’’(v(\chi)) = f’’(v(x)) = 0$, and so $f(\chi) = u’(\lambda)$ for some $\lambda \in N’$. Now $d(x’’)$ can be written as the image of $\lambda$ in $\text{Coker}(f’)$, so $d(x’’) = [\lambda] = \lambda + \text{Im}(f’)$. So I need to show that $y’ - \lambda \in \text{Im}(f’)$.
But my next calculations shows the following: since $v(x - \chi) = v(x) - v(\chi) = 0$, and since $u$ is injective, we have $u(0) = x - \chi$. And so:
\begin{alignat}{2} &\Rightarrow\quad &f(u(0)) &= u’(f’(0))\\ &\Rightarrow &f(x - \chi) &= 0\\ &\Rightarrow &f(x) - f(\chi) &= 0\\ &\Rightarrow &u’(y’) - u’(\lambda) &= 0\\ &\Rightarrow &u’(y’ - \lambda) &= 0\\ &\Rightarrow &y’ - \lambda &= 0\\ \end{alignat}
So in fact I get more than what I hoped for. Is there anything wrong with my calculations?
As is written in the comments, the injectivity of $u$ doesn't gives us that $$v(x-\chi)=0 \implies u(0)=x-\chi.$$
If there is a $\chi \in M$ with $v(\chi)=x''=v(x)$, then $x-\chi \in \ker(v) = \operatorname{im}(u)$, and so exists a $x' \in M'$ such that $u(x') = x-\chi$. If, as you did, we find some $\lambda \in N'$ with $f(\chi) = u'(\lambda)$, it follows that $$\begin{align*} u'(y'-\lambda) &= u'(y')-u'(\lambda) \\ &= f(x)-f(\chi) \\ &= f(x-\chi) \\ &= f(u(x')) = u'(f'(x')) \end{align*}$$ and then $y'-\lambda = f'(x') \in \operatorname{im}(f')$ because $u'$ is injective. Hence $y'$ and $\lambda$ have the same image in $\operatorname{coker}(f')$.