According to Wikipedia, the image of a morphism $f:A \to B$ is a monomorphism $m:I \to B$ for which there is some $e:A \to I$ with $f=m\circ e$, and such that the following universal property is verified: if there is another monomorphism $m':I' \to B$ for which there is an $e':A \to I'$ such that $f=m' \circ e'$, then, there is a unique morphism $h:I \to I'$ making the following diagram commute:
Now, I tried applying this definition to the category of sets:
Let $f:A \to B$ be a function. The obvious choices are:
- $I=\text{Im}(f)$
- $e$ is the (codomain) restriction of $f$ to its image
- $m$ is the inclusion of the image of $f$ to $B$.
The problem arises when I want to find a choice of $h$. Since it has to be $h(e(a))=e'(a)$, it is $h(f(a))=e'(a)$. But then, this is not necessarily well defined, because it could be the case that there are $a,a'\in A$ such that $f(a)=f(a')$ but $e'(a) \neq e'(a')$, right?
What am I missing here? Could it be the case that Set does not admit images in the categorical sense? That would be so unintuitive.
Edit: your comment is right - as NDB notes though, $f(a)=f(a')$ implies $e'(a) = e'(a')$ by commutativity and since $m'$ is mono, so this issue doesn't come up.
You are correct that the commutativity of the lower left triangle implies that we must have $h(e(a)) = e'(a)$. Note that your choice of $e$ is surjective, so this determines $h$ uniquely. So it's just left to check commutativity of the triangles. For the lower right one you can use surjectivity of $e$ and commutativity of the other (upper, lower left and outer) triangles.