# Universal property of images in category theory

Let $$\mathcal{A}$$ be an additive category with all kernels and cokernels and $$f:A\to B$$ a morphism. If $$e:B\to \text{coker}(f)$$ is the canonical epimorphism, define $$\text{im}(f):=\ker(e)$$, with a canonical monomorphism $$i:\text{im}(f)\to B$$. Prove that:

$$1)$$ There is a unique $$\pi:A\to\text{im}(f)$$ such that $$i\circ\pi=f$$

$$2)$$ If there is a monomorphism $$i':C\to B$$ and a morphism $$\pi':A\to C$$ such that $$i'\circ\pi'=f$$, then there is a unique morphism $$\mu:\text{im}(f)\to C$$ such that $$\mu\circ\pi=\pi'$$ and $$i'\circ\mu=i$$.

For part $$1)$$, I used the fact that $$e\circ f=0$$ (by definition of $$\text{coker(f)}$$), so by the universal property of $$\ker(e)$$, there is a unique $$\pi$$ such that $$i\circ\pi=f$$

For $$2)$$, I've shown that if there is another $$\mu'$$ with these properties, then $$i'\circ \mu=i=i'\circ \mu'$$ and, since $$i'$$ is a monomorphism, then $$\mu'=\mu$$, so $$\mu$$ is unique. Furthermore, assuming $$i'\circ\mu=i$$, we get $$i'\circ\mu\circ\pi=i\circ\pi=f=i'\circ\pi'$$ and, since $$i'$$ is a monomorphism, $$\mu\circ\pi=\pi'$$, which means we only need to find $$\mu$$ with $$i'\circ\mu=i$$. Here is where I'm stuck, because I don't know how to come up with an arrow $$\textit{leaving }\text{im}(f)$$, since the universal property of $$\ker(e)$$ can only give an arrow $$\textit{arriving}$$ at it.

• If $\alpha:B\to C$ is a monomorphism, composable after $\def\fii{\varphi} \fii:A\to B$, then $\def\im{\rm im} \im(\alpha\circ \fii)\cong\im(\fii)$ should hold. For start, prove that the pushout of $B\to{\rm coker}(\fii)$ along $\alpha$ is $C\to{\rm coker}(\alpha\circ\fii)$. Apr 3 '17 at 23:34
• 1) The proof is not correct. You cannot argue with $\pi$ before defining it. The claim follows directly from $e f=0$ and the universal property of $i$. Apr 4 '17 at 19:06
• @HeinrichD, of course you're right, silly mistake. Still trying to figure 2) though Apr 4 '17 at 21:51
• You've tagged this question "abelian-categories", but your hypothesis is that $\mathcal{A}$ is only additive. Which one do you really want? Apr 6 '17 at 15:26
• @ArnaudD., the tag was misleading indeed. The hypothesis is just $\mathcal{A}$ additive with all kernels and cokernels Apr 6 '17 at 16:02

This is not true in general. For instance, let $\mathcal{A}$ be the category of torsion-free abelian groups. This is an additive category with kernels and cokernels (to form a cokernel, first take the cokernel in $Ab$ and then mod out the torsion subgroup). Now consider the map $f:\mathbb{Z}\to\mathbb{Z}$ given by multiplication by $2$. The cokernel of $f$ is $0$, so the image of $f$ is the identity $\mathbb{Z}\to\mathbb{Z}$. But taking $i'=f$ and $\pi'=1$, $i'$ is a monomorphism, $i'\circ\pi'=f$, but $i=1$ does not factor through $i'$.