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This is an exercise from Rotman , Introduction to homological algebra.

Given a pushout diagram in $R$-Mod

$$\begin{array} AA & \stackrel{g}{\longrightarrow} & C \\ \downarrow{f} & & \downarrow{\beta} \\ B & \stackrel{\alpha}{\longrightarrow} & D \end{array} $$

prove that $g$ injective implies $\alpha$ injective, and that $g$ surjective implies $\alpha$ surjective.

I have problems with the injective part, how to solve it ?

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    $\begingroup$ You can either do a concrete calculation (because you are working with modules), or you can make an abstract argument (because this is true for all abelian categories). $\endgroup$
    – Zhen Lin
    Jun 1, 2014 at 6:43
  • $\begingroup$ @ZhenLin: thank you Zhen, but I don't know how to start; can you give me some concrete hints ? $\endgroup$
    – WLOG
    Jun 1, 2014 at 9:31
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    $\begingroup$ Do you know what a pushout in the category of $R$-modules is, concretely? If you do, just calculate! $\endgroup$
    – Zhen Lin
    Jun 1, 2014 at 10:44
  • $\begingroup$ @ZhenLin: I wrote an answer $\endgroup$
    – WLOG
    Jun 1, 2014 at 11:50

1 Answer 1

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So I applied the construction of pushout in $R$-Mod: $$D \cong (B \oplus C)/S$$ where $S$ is the submodule generated by $$\lbrace (f(a), -g(a) ) | \ a \in A \rbrace$$

Thus if $\alpha(b) = 0 $ we have that $$\alpha(b) = (b,0) \in S \Rightarrow \exists a \in A \ \text{s.t} \ \ (b,0) = (f(a) , -g(a))$$

But this implies $a= 0$ due to injectivity of $g$ and then $b= 0$.

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