# Image in Abelian Categories and Relation to Image of Function

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 $$\phi$$.

• 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. – Eric Wofsey Feb 9 '20 at 5:04
• 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 $\in$. – Aaron Feb 9 '20 at 5:22

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 $$\alpha$$.
• 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 $i$? – Aaron Feb 9 '20 at 21:44
• You mean from the domain of $\phi$? Then yes, if you require additionally that $ig=f\phi$. – Eric Wofsey Feb 9 '20 at 22:13