3
$\begingroup$

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 ?

$\endgroup$
  • 2
    $\begingroup$ 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 '14 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 '14 at 9:31
  • 1
    $\begingroup$ Do you know what a pushout in the category of $R$-modules is, concretely? If you do, just calculate! $\endgroup$ – Zhen Lin Jun 1 '14 at 10:44
  • $\begingroup$ @ZhenLin: I wrote an answer $\endgroup$ – WLOG Jun 1 '14 at 11:50
4
$\begingroup$

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 $b= 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.