# Isomorphism theorem in categories?

In group theory, we know that a group homomorphism $f:G \to H$ induces an isomorphism of groups ${G \over \ker(f)} \simeq \operatorname{im}(f)$. I've searched a lot for a categorical version of this theorem in the following sense but I cannot find anything.

Let $F:\mathcal{A \to B}$ be a functor between abelian categories. Does this functor induce an isomorphism of categories ${\mathcal{A} \over \mathcal{Ker}(F)} \simeq \mathcal{B}$? (The quotient here is serre quotient as I mentioned in commnet to answer by Max)

Here, by $\mathcal{Ker}(F)$ I mean the full subcategory of $\mathcal{A}$ consisting of objects $A$ so that $F(A)=0$.

• It's rather $G/\ker (f) \simeq \mathrm{im} f$. – Max Aug 12 '18 at 8:16

I don't know what you mean by quotienting by a subcategory, but here's an idea:

if $F: C\to D$ is a functor bewteen arbitrary categories, then you have a relation on morphisms (which we might denote $\ker(F)$) that is, for $f,g: A\to B$, $(f,g)\in\ker(F)\iff F(f)=F(g)$ (note that we do ask that $f,g$ have the same domain and codomain).

This relation is very clearly a congruence on $C$, that is : if $f\sim f', g\sim g'$, then $f\circ g \sim f'\circ g'$. This allows us to form the quotient category $C/\ker (F)$ whose objects are those of $C$ and morphisms are equivalence classes of morphisms.

Together with this we have a canonical functor $\Pi: C\to C/\ker(F)$ that satisfies the universal property you probably expect (note that you can do the same thing with any congruence, which shows that congruences and "kernels" are the same; just as for groups)

Now what's interesting here is that it shows that $F$ can be factored as $\tilde{F}\circ \Pi$ where $\tilde{F}$ is faithful, and $\Pi$ is full and surjective on objects, which should remind you of similar factorizations in various categories.

In particular, $\tilde{F}: C/\ker(F) \to D$ is faithful and so makes $C/\ker(F)$ equivalent to the essential image of $F$ (but of course we can't expect $\tilde{F}$ to be full if $F$ wasn't ! However if $F$ is full, and essentially surjective, then we do get that $C/\ker(F)$ is equivalent to $D$)

Until here, no need for abelian categories. What changes a bit in abelian categories is essentially the same thing as what happens when you go from sets to groups: you can exchange your congruence for a normal subgroup (or here indeed a subgroup). That is, if you have an additive functor $F:C\to D$, then instead of having a congruence you can have, for each $A,B\in \mathrm{Ob}C$ a subgroup of $\hom(A,B)$ (the subgroup of those $f$ such that $F(f) = 0$: it is indeed a subgroup if $F$ is additive), and you can again construct $C/\ker(F)$; but you still won't have an equivalence with $D$ unless $F$ is full and essentially surjective to begin with.

I think (though I haven't checked that thoroughly) that $C/\ker(F)$ will be abelian as well and $\Pi$ additive.

I can somehow relate this to quotienting by the subcategory of the $A$'s with $F(A) = 0$ in the following sense. If $F(A) = 0$, then for any $B$ and $f,g: B\to A$, $F(f)=F(g)= 0_{F(B),F(A)}$ and so $f\sim g$ and so in $C/\ker(F)$, $A$ is terminal (and similarly initial) so it's isomorphic to $0$: in a sense you've turned $A$ into $0$ (perhaps that's what you meant by quotienting by it)

• Thanks for your time. Here, the quotient is (serre) quotient category as defined in [link] (arxiv.org/pdf/1210.1425.pdf). In this paper, I finally understand when the functor F induces such equivalence. – math16 Aug 12 '18 at 8:54
• Ah, sorry then I completely missed the point ! You should probably add some clarification in your question so that future answers may be more on point ! (however it's not clear to me why your subcategory $\ker(F)$ is thick in the sense defined in the article) – Max Aug 12 '18 at 9:21
• You may use the stronger version of definition as in ncatlab.org/nlab/show/thick+subcategory said. – math16 Aug 12 '18 at 9:57
• I still don't see it, but it must be me (why, if $B$ is a subobject of $A$ and $F(A)=0$, must $F(B)=0$ ? ) – Max Aug 12 '18 at 10:12
• Oh, $F$ is assumed to be exact, ok ! you should probably mention that in your question as well – Max Aug 12 '18 at 10:28