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I am trying to understand the notion of image when extended to abelian categories. In an abelian category, for a morphism $f$ we have $\operatorname{Im} f = \operatorname{Ker} \operatorname{Coker} f$.

A kernel of $f$ is a morphism $g$ where $fg = 0$ and $f\sigma = 0 \Rightarrow \sigma = g\sigma_0$ for unique $\sigma$. I write $g \in \operatorname{Ker} f$ if $g$ is a kernel of $f$. I have suppressed the source and target categories for the morphisms. The definitions are analogous for cokernels.

Is it true that for any morphism $\phi$, we have $f\phi \in \operatorname{Im}f$, just as for some element $x$ we have $f(x) \in \operatorname{Im}f$ when $f$ is a function?

I have tried to prove this, but cannot seem to prove that $f\phi$ obeys the universal property of kernels. That is, for $\alpha \in \operatorname{Coker}f$, obtaining $\alpha f \phi = 0$ is clear but I can't show that $\alpha \sigma = 0 \Rightarrow \sigma = f\phi \sigma_0$ for a unique $\sigma_0$.

Since $\alpha$ is a cokernel of $f$, we have that $\sigma = f\tilde{\sigma}_0$ for some unique $\tilde{\sigma}_0$, but I am stuck here and do not know how to construct the map with $\phi$.

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  • $\begingroup$ What is $\phi$? What do you mean by $f\phi \in \operatorname{Im}f$? You seem to be using the symbol $\in$ in a rather nonstandard way. $\endgroup$ – Eric Wofsey Feb 9 '20 at 5:04
  • $\begingroup$ I've edited the question to hopefully be clearer. $\phi$ is a morphism. I understand that kernel, cokernel, and image are universal constructions so that we can speak of "the kernel, etc" up to isomorphism, but I want to be more pedantic and think of it as the set of all morphisms satisfying the necessary conditions, hence the usage of $\in$. $\endgroup$ – Aaron Feb 9 '20 at 5:22
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No, this is horribly wrong. Let's suppose your category is the category of modules over some ring. Then if $f:M\to N$ is a morphism, the inclusion map $i:f(M)\to N$ of the ordinary set-theoretic image of $f$ is an image of $f$ in the categorical sense. What you are then asserting is that for any morphism $\phi:K\to M$ for any other module $K$, the composition $f\phi$ is another image of $f$. This is obviously false, because it would in particular mean that $K$ must be isomorphic to $f(M)$ (since images are unique up to isomorphism), but $K$ could be any module at all (and $\phi$ could be the zero map).

The correct analogous statement is that $f\phi$ factors through $\operatorname{Im} f$. That is, if $i$ is an image of $f$, then there exists a morphism $g$ such that $f\phi=ig$ (and moreover this $g$ is unique since $i$ is a kernel and thus monic). Indeed, this is immediate from the universal property: if $\alpha$ is a cokernel of $f$ then $\alpha f\phi=0\phi=0$, so $f\phi$ factors uniquely through $i$ since it is a kernel of $\alpha$.

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  • $\begingroup$ I see, thanks for the clear answer. So is correct to say that the categorical version of the statement "$f(x) \in \operatorname{Im} f$ for a function $f$" is that for $f\phi$ there exists a morphism $g$ from the kernel object of $\phi$ to the "image object" of $i$? $\endgroup$ – Aaron Feb 9 '20 at 21:44
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    $\begingroup$ You mean from the domain of $\phi$? Then yes, if you require additionally that $ig=f\phi$. $\endgroup$ – Eric Wofsey Feb 9 '20 at 22:13

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