Given the following diagram, I want to show that if $f$ is injective, then $\alpha$ is injective, and then if $f$ is surjective, $\alpha$ is surjective.

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

Before I begin to prove either of these statements, I want to make sure I am defining each module and map correctly. I believe I want $X=(B\oplus C)/S$, where $S=\langle f(b),-g(c)\rangle$, and $\alpha$ and $\beta$ to be projection maps, where $\alpha((b,c)+S)=c$, and $\beta((b,c)+S)=b$.

For my proof, here is my idea:

I ultimately want to prove that if $\alpha((b,c)+S)=0$, this implies $c=0$. So I choose some $(b_1,c_1)+S\in X$ such that $\alpha((b_1,c_1)+S)=0$. Then, $(b_1,c_1)\in\ker(\alpha)$. At this point, I think I want to be able to say something about $(b_1,c_1)$ using the equivalence relation $f(b)=g(c)$, but I am not sure how to go about this. Any suggestions would be helpful!

  • 1
    $\begingroup$ The thing you are describing would be a pushout, if $f, g$ would have the opposite direction that they have (note that this way, your definition of $S$ is invalid: just check domains and codomains). A construction of a pullback is rather $\{(b, c) \in B \times C\;|\; f(b)=g(c)\}\subseteq B \times C$. $\endgroup$ – Pavel Čoupek May 4 '18 at 2:46
  • $\begingroup$ Okay thank you for the clarification. So this is the way $S$ should be defined instead of how I have defined it, is it still true that $X=(B\oplus C)/S$, or should it be $A$ instead? Also, are $\alpha$ and $\beta$ defined correctly? $\endgroup$ – MathStudent1324 May 4 '18 at 3:29
  • $\begingroup$ The pullback will not be a quotient of $B\oplus C$ by any $S$. It will be the submodule of $B \times C(=B \oplus C)$. The pullback can be taken as $\{(b, c)\in B \times C\;|\; f(b)=g(c)\}$, and $\alpha, beta$ will be projections to the second component, first component respectively (so in that sense yes, the $\alpha$ and $\beta$ are induced by projections). $\endgroup$ – Pavel Čoupek May 4 '18 at 3:37

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