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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 $\psi(A)$.

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I was reluctant to use the book, but ended up resorting to it. Here's my understanding of the proof.

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.

There was just more usefulness for the diagram that I didn't see!

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  • $\begingroup$ why we suppose $\beta (b) = 0$? $\endgroup$ – Emptymind Oct 2 '17 at 5:21
  • $\begingroup$ "$\phi (b)=0$ by injectivity of $\gamma$" I do not understand this statement, could you say it in details please? $\endgroup$ – Emptymind Oct 2 '17 at 5:53

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