Generalization of set theoretic image, viewed as a function?

120 Views Asked by At

Given a function $f:A \to B$, we can use image of the function to create $F:\mathcal{P}A \to \mathcal{P}B$, defined as $F(X) = f(X)$. How could one generalize this to category theory with power objects? Given a morphism $g:C \to D$, can we find another morphism $G:\Omega^C \to \Omega^D$, which would be analogous to the set theoretic one given above?

Image is generalized as the smallest subobject of a codomain throught which a function factors. I was thinking about using a subobject classifier and somehow find a morphism that would help me, but I couldn't think of anything to connect $\Omega^C$ and $\Omega^D$ in a meaningful way.

1

There are 1 best solutions below

3
On

Actually there are multiple different generalizations of the image available in category theory. In any category with finite limits and colimits you can define the regular image and regular coimage; the regular image of a morphism $f : x \to y$ is the equalizer of its cokernel pair, and dually the regular coimage is the coequalizer of its kernel pair. These are generalizations of "kernel of the cokernel" and "cokernel of the kernel" respectively to "nonlinear" settings. Every morphism then factors as

$$x \to \text{coim}(f) \to \text{im}(f) \to y$$

where the morphism $x \to \text{coim}(f)$ is a regular epimorphism, the morphism $\text{im}(f) \to y$ is a regular monomorphism, and the morphism $\text{coim}(f) \to f$ is both an epimorphism and a monomorphism. If the ambient category is balanced, meaning that every map which is both epi and mono is an isomorphism, then $\text{coim}(f) \to \text{im}(f)$ is an isomorphism and the regular image and coimage coincide; this happens in particular in any abelian category. But in general this won't be true, and the regular image and coimage capture two genuinely different notions of "image."

Here are two suggestive examples:

  1. In the category of commutative rings, if $f : R \to S^{-1} R$ is any localization of rings, then the regular coimage is $R$ but the regular image is $S^{-1} R$.

  2. In the category of topological spaces, if $f : X \to Y$ is a continuous function, then the regular coimage is the set-theoretic image topologized with the quotient topology from $X$, while the regular image is the set-theoretic image topologized with the subspace topology from $Y$.

So, generally and loosely speaking, the regular coimage is "the image as seen from the perspective of $X$," while the regular image is "the image as seen from the perspective of $Y$."

One can define more generally an image corresponding to any class of monomorphisms you like, as the smallest monomorphism in that class through which a given morphism factors. But there's no guarantee that such a thing will exist in general, whereas the regular image and coimage always exist and are relatively easy to compute under mild hypotheses.