# How do you prove a piece of the Short Five Lemma?

Let $\alpha, \beta, \gamma$ be a homomorphism of short exact sequences, in that order. Then if $\alpha, \gamma$ are injective, then so is $\beta$.

Let the sequeces be: $$\begin{matrix} 0 & \to & A & \xrightarrow{\psi} & B & \xrightarrow{\phi} & C & \to & 0 \\ \ & \ & \downarrow^{\alpha} & \ & \downarrow^{\beta} \ & \ & \downarrow^{\gamma} \\ 0 & \to & A' & \xrightarrow{\psi'} & B' & \xrightarrow{\phi'} & C' & \to & 0 \end{matrix}$$

Then since $\psi, \psi', \alpha$ are injective and $\beta \psi = \psi' \alpha$, we have that $\beta$ is injective on $\psi(A)$.

Suppose $\beta(b) = 0$. Then $\gamma \phi(b) = \phi'\beta(b) = \phi'(0) = 0$, so $\phi(b) = 0$ by injectivity of $\gamma$. Then $\ker \beta \subset \ker \phi = \psi(A)$. So our $b = \psi(a)$ for some $a \in A$. By commuting diagrams again $\beta \psi (a) = 0 =\psi' \alpha (a)$, and by injectivity $a = 0$, so $b = \psi(0) = 0$. We're done.
• why we suppose $\beta (b) = 0$? – Emptymind Oct 2 '17 at 5:21
• "$\phi (b)=0$ by injectivity of $\gamma$" I do not understand this statement, could you say it in details please? – Emptymind Oct 2 '17 at 5:53