pullback of kernel is kernel

Question

I'm trying to prove Prop. 37, part (iii) of these notes on abelian categories (where it is left as an exercise): http://therisingsea.org/notes/AbelianCategories.pdf.

Part $(i)$ follows readily from the universal property of the pullback and part $(ii)$ is proved in the linked PDF, but I'm not sure how to do part $(iii)$.

Answer 1

Use the universal property of the pullback to prove that of the kernel.

More specifically, suppose $f:X\to C_2$ is such that $\beta \alpha_2 f=0$; then $\alpha_2f$ factors through $\alpha_1$, in the sense that $\alpha_2f=\alpha_1 g$ for some $g$. Then, by the universal property of the pullback, there must be some $h$ such that $f=\pi_2h$. Uniqueness of such an $h$ follows from $(i)$.

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Alternatively, you can show this by pasting pullback diagrams : $\require{AMScd}$ \begin{CD} P @>{\pi_2}>> C_2\\ @V{\pi_1}VV @VV{\alpha_2}V\\ C_1 @>>{\alpha_1}> C \\ @V{}VV @VV{\beta}V\\ 0 @>>{}> D \end{CD} The top square is a pullback, and by definition of kernel, so is the second one; hence the rectangle is also a pullback, and $\pi_2$ is the kernel of $\beta\alpha_2$.