Suppose we have a homomorphism $\alpha, \beta, \gamma$ of short exact sequences: $$ \begin{matrix} 0 & \to & A & \xrightarrow{\psi} & B & \xrightarrow{\phi} & C & \to & 0 \\ \ & \ & \downarrow^{\alpha} & \ & \downarrow^{\beta} \ & \ & \downarrow^{\gamma} \\ 0 & \to & A' & \xrightarrow{\psi'} & B' & \xrightarrow{\phi'} & C' & \to & 0 \end{matrix} $$

If both $\alpha, \gamma$ are surjective then so is $\beta$. This can be proved using the properties of the diagram somehow.

I've tried several things.


It should be straightforward.

We want to prove $\beta$ is surjective, so start out from an arbitrary element $b'\in B'$. We can do one thing: consider $c':=\phi'(b')\in C'$.
Since $\gamma$ is surjective, we get $c$ with $\gamma(c)=c'$.

That the pair $\phi,0$ of maps is exact means nothing else but that $\phi$ is surjective. It yields an element $b\in B$, such that $\gamma\phi(b)=c'=\phi'(b')$.

Now we might not get $\beta(b)=b'$ with this element $b$, nevertheless, we have that $\beta(b)$ and $b'$ has the same image under $\phi'$, so by exactness, this gives $a'\in A'$ such that $\psi'(a')=b' - \beta(b)$.

Can you take it on from here?

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  • $\begingroup$ "It should be straightforward." Does that introductory sentence serve any useful purpose? $\endgroup$ – Junglemath Jul 11 at 18:14
  • $\begingroup$ Yes. When doing diagram chasing, usually there's a unique straightforward flow of the proof. $\endgroup$ – Berci Jul 11 at 20:12

Thanks to Berci and Pedro Tamaroff.

Let $b' \in B'$. Then $\phi'(b') = c' \in C'$ and by surjectivity of $\gamma$, there's $c \in C$ with $\gamma(c) = \phi'(b')$ and there's $b \in B$ with $\phi(b) = c$. So $\gamma\phi(b) = \phi'(b') = \phi'(\beta(b))$. So $\beta(b) - b' \in \ker \phi' = \psi'(A') = \psi' \alpha(A)$. So there's $a \in A$ so that $\psi'\alpha(a) = \beta(b) - b' = \beta \psi(a)$. Then clearly $b'$ can be written as the image of the element $b - \psi(a)$ under $\beta$. We're done.

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