Suppose that we have a Long Exact sequence
$$…\rightarrow A \xrightarrow{a} B \xrightarrow{b} C \xrightarrow{c} D \xrightarrow{d} E \xrightarrow{e} F \rightarrow…$$
$\DeclareMathOperator{\im}{im}\DeclareMathOperator{\coker}{coker}$
Show then that the following induced short sequence is exact:
$0 \rightarrow\coker(a) \xrightarrow{\tilde{b}} C \xrightarrow{\tilde{c}} \im(c) \rightarrow 0$
Proof:
Let $\bar{x} \in B/\im(a) = \coker(a)$ and suppose $\tilde{b}(\bar{x})=0$
Then $b(x)=0$
So $x \in \ker(b) = \im(a)$
so $\bar{x} \in \im(a)$
$\rightarrow \bar{x} = 0$
Thus $\tilde{b}$ is injective and thus the S.E.S. is exact at $\coker(a)$
I will now show exactness at $C$.
let $x \in C$ and $\bar{c}(x)=0$
This implies that $c(x)=0$
so $x \in \ker(c) = \im(b)$
so $b(y)=x$ for some $y \in B$.
Thus $\bar{b}(\bar{y})=\bar{x}$
And so $\ker(\bar{c}) \subset \im(\bar{b})$
Now Let $x \in C$ and suppose that $\bar{b}(\bar{y})=x$
then $x \in \im(\bar{b})$
so $x \in \im(b) = \ker(c)$
thus $\bar{x} \in \ker(c)$
Thus this S.E.S. is exact at C.
Is this correct so far?
Best Answer
Let $\tilde c : C \to \text{im} C$ be the restriction of $c : C \to D$ to its image. Let $\tilde b : \text{coker} a \to C$ be the induced map on the quotient. That is, if $q : B \to \text{coker} a$ is the quotient map then $\tilde b$ satisfies $b = \tilde b \circ q$
It is clear from the definition of $\tilde c$ that $\tilde c$ is surjective.
Note that $\text{coker} a = B / \text{im}a = B / \text{ker}b$ by exactness. So $\tilde b$ is injective.
Lastly if $x \in \ker \tilde c$ then $x \in \ker c$ and by exactness $x \in \text{im} b$. From the definition above it is easy to see that $\text{im} b = \text{im} \tilde b$, so we are done.