Cokernel of a column map between objects in abelian category

Question

Let $\mathcal{A}$ be an abelian category, let $f\colon X \to X_1$ and $g \colon X \to X_2$, take the map to the direct sum $\binom{f}{g} \colon X \to X_1 \oplus X_2$. What can I say about the image, and then about the cokernel, of the map $\binom{f}{g}$ in terms of image and cokernel of $f$ and $g$?

Answer 1

The cokernel of $\begin{pmatrix}f \\ g\end{pmatrix}\colon X\to X_1\oplus X_2$ is not really anything in terms of $\operatorname{coker} f$ and $\operatorname{coker} g$. Consider for example $X_1=X_2=X$ and $f = g = \mathrm{id}_X$. Then $\begin{pmatrix}\mathrm{id}_X \\ \mathrm{id}_X\end{pmatrix}\colon X\to X\oplus X$ is the diagonal map, which has nontrivial cokernel, while the cokernel of $\mathrm{id}_X$ is trivial.

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Intuitively, $\operatorname{coker} \begin{pmatrix}f \\ g\end{pmatrix}$ is the quotient of $X_1\oplus X_2$ by $$\tag{*} \{ (f (x), g (x)) \mid x\in X \}.$$ You may consider $$\operatorname{coker} (f\oplus g\colon X\oplus X\to X_1\oplus X_2) \cong \operatorname{coker} f\oplus \operatorname{coker} g,$$ and it will be the quotient of $X_1\oplus X_2$ by $$\tag{**} \{ (f (x), g (y)) \mid x,y\in X \}.$$ Note that (*) is "smaller" than (**), so $\operatorname{coker} \begin{pmatrix}f \\ g\end{pmatrix}$ is "bigger" than $\operatorname{coker} (f\oplus g)$.

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Probably the best thing one can do is to observe that the composition $$X\xrightarrow{\begin{pmatrix}f \\ g\end{pmatrix}} X_1\oplus X_2\twoheadrightarrow \operatorname{coker} (f\oplus g)$$ is zero, so by the universal property of cokernels, the canonical morphism $X_1\oplus X_2\twoheadrightarrow \operatorname{coker} (f\oplus g)$ should factor through $\operatorname{coker} \begin{pmatrix}f \\ g\end{pmatrix}$: $$X_1\oplus X_2\twoheadrightarrow \operatorname{coker} \begin{pmatrix}f \\ g\end{pmatrix} \xrightarrow{\exists !} \operatorname{coker} (f\oplus g)$$ So there is a canonical epimorphism from $\operatorname{coker} \begin{pmatrix}f \\ g\end{pmatrix}$ to $\operatorname{coker} (f\oplus g)$, which justifies our intuitive observation above.

You can also see that the cokernel of $\begin{pmatrix}f \\ g\end{pmatrix}$ is given by $X_1\oplus X_2 \xrightarrow{(\overline{g}, -\overline{f})} X_1\sqcup_X X_2$, where $\overline{g}\colon X_1\to X_1\sqcup_X X_2$ and $\overline{f}\colon X_2\to X_1\sqcup_X X_2$ are the canonical arrows coming from the pushout square. But this is very tautological.