# Should the isomorphism theorems be seen as an "interface" between algebra and category theory?

My first instinct when I thought about algebra in category theory, was to try to "generalize the isomorphism theorems in category theory".

So I tried to prove the generalization of "the image of a group homomorphism is isomorphic to the quotient group generated by its kernel".

But then I found out that in category subobjects are actually defined in terms of monomorphisms, which for the category Grp is essentially implicitly using that isomorphism theorem.

• So is it correct that I shouldn't be trying to prove the isomorphism theorems in category theory?

• Is it correct that instead, the isomorphism theorems should be seen as justifying talking about algebraic structures (among other structuers) in terms of structure preserving morphisms? in that sense they are like the "interface" between category theoretical algebra (e.g. talking about groups in terms of group homomorphisms) and "set-theoretic" algebra (talking about groups in terms of the elements of the group, and cosets and so forth).

## 2 Answers

Before addressing your questions I will write here my favorite version of the first isomorphism theorem. As others have commented, you need notions of quotients, images and kernels before even attempting to enunciate such a result. There's more than one way to do this (for example, additive categories). Here we are going to work with concrete categories. Recall the notion of images.

Definition [kernel]: Let $$f : X \to Y$$ be a function. The kernel of $$f$$ is the set $$\{(a,b) \in X \times X \mid f(a)=f(b)\}$$.

The notion of kernel as defined here is simply the kernel pair of $$f$$, that is, the pullback of $$X \xrightarrow{f} Y \xleftarrow{f} X$$.

Definition [concrete quotients and congruences]: Let $$(C,U)$$ be a concrete category and $$X$$ an object of $$C$$. A concrete quotient of $$X$$ is an epimorphism $$\pi : X \to Y$$ such that $$U(\pi)$$ is epi and for every object $$Z$$ of $$C$$ and every function $$f : U(Y) \to U(Z)$$, the following are equivalents:

• There exists a morphism $$f' : Y \to Z$$ such that $$U(f') = f$$.
• There exists a morphism $$g : X \to Z$$ such that $$U(g) = f \circ U(\pi)$$.

The set $$\ker(U(\pi))$$ is called a congruence on $$X$$.

If you prefer, you can define concrete quotients as equivalence classes instead. Note that this notion of quotient coincides with topological quotients, for instance, while the usual notion of quotients (that is, epimorphisms) does not. In essence, concrete quotients allow you to complete diagrams in the base category by looking at the underlying diagrams in $$Set$$. A congruence on an object $$X$$ is essentially an equivalence relation on $$U(X)$$ with an associated concrete quotient of $$X$$. Observe, however, that congruences need not to arise only from $$U(\pi)$$ for $$\pi$$ a concrete quotient.

Theorem [the First Isomorphism Theorem]: Let $$(C,U)$$ be a concrete category, where $$C$$ is complete and $$U$$ is continuous. Let $$q : X \to Z$$ be a morphism in $$C$$ such that $$\ker(U(q))$$ is a congruence on $$X$$. Then the morphism $$m : X/\ker(U(q)) \to Z$$ (such that $$q = m \circ \pi_q$$) is the image of $$q$$.

Proof : First of all we must verify that $$m$$ is a monomorphism. Let $$x,y \in U(X)$$ and $$[x],[y]$$ their equivalence classes regarding $$\ker(U(q))$$. If $$U(m)([x])=U(m)([y])$$, then $$(U(\pi_q) \circ U(m))(x)=(U(\pi_q) \circ U(m))(y)$$, hence $$U(\pi_q \circ m)(x)=U(\pi_q \circ m)(y)$$, which implies $$U(q)(x)=U(q)(y)$$. Therefore $$(x,y) \in \ker(U(q))$$ and $$[x]=[y]$$. $$U(m)$$ is mono, hence $$m$$ is as well ($$U$$ is faithful).

Now let $$m' : Y \to Z$$ be a monomorphism and $$h : X \to Y$$ be a morphism such that $$q= m' \circ h$$. we wish to prove the existence of $$f : X/\ker((U(q))) \to Y$$ such that $$m = m' \circ f$$. If $$(x,y) \in \ker(U(q))$$, then $$U(q)(x)=U(q)(y)$$, hence $$U(h)(x)=U(h)(y)$$ (since $$U$$ is continuous and $$m'$$ is mono, $$U(m')$$ is mono). By the definition of concrete quotients, there exists a morphism $$f : X/\ker((U(q))) \to Y$$ such that $$h = f \circ \pi_q$$. Since $$m' \circ h = q = m \circ \pi_q$$, we have $$m' \circ f \circ \pi_q = m \circ \pi_q$$. Since $$\pi_q$$ is epi, we have $$m' \circ f = m$$.

Note that, in particular, this isomorphism theorem is valid on the category of topological spaces (with the obvious forgetful functor to $$Set$$)! What is the problem here? I'll leave that as an exercise.

Now, addressing the questions:

So is it correct that I shouldn't be trying to prove the isomorphism theorems in category theory?

That is not correct. However, you should be aware of the fact that general categories might not always have the structure/properties you need to talk about certain concepts. In that case, however, you should still be able to consider a particular class of categories in which you can prove your desired results.

Is it correct that instead, the isomorphism theorems are kind of like "interfaces", which justify talking about algebraic structures (among other structuers) in terms of structure preserving morphisms?

I'm not sure if I understand this question. The fact that homomorphisms between algebraic structures satisfy the isomorphism theorems is certainly a good reason to talk about structure preserving functions (instead of non-structure preserving functions) in certain scenarios. However, in other structures, where the theorem might not be valid (topological spaces for example), it is still "better" to consider structure preserving functions than simply general functions.

• I wish I had known this version of the isomorphism theorem before writing my answer, +1 Commented Apr 5, 2019 at 17:47
• For the LaTeX, writing f:X\to Y ($f:X\to Y$) as opposed to f\colon X\to Y ($f\colon X\to Y$) seems easier and renders better... (at least for me). Commented Apr 6, 2019 at 1:27
• @Hilario Very nice answer! Could you share where did you find this result? Or did you come up with it? Commented Nov 9, 2023 at 19:28
• @FernandoChu I did in fact come up with the result, but the fact I wasn't able to find it in the literature likely means that either 1) I didn't look hard enough or 2) The result is folklore/no very interesting. Commented Nov 20, 2023 at 23:41

This isn't a full answer as I don't understand half of the question, and have asked for precision on the other half, but it's too long to be a comment

No, it's not implicitly using the isomorphism theorem, it's using the fact that (in algebraic structures) the corestriction of an injective morphism to its image is an isomorphism, which is way more basic than the first isomorphism theorem.

Then, for your questions :

$$\bullet$$ No you shouldn't try to prove the isomorphism theorem in general categories because it simply isn't true in general. First of all, you would have to have a notion of image and of kernel, which don't usually make sense in an arbitrary category, and even when they do exist, it's not true that the theorem holds. For me to make a precise statement and give counterexamples here you have to tell me what you mean by "image" in a general category, for instance are you referring to this definition ?

$$\bullet$$ I don't understand this question. Let me just say how I feel about the isomorphism theorem (the first one, the others are just immediate corollaries) for groups, and algebraic structures more generally, in the hope that it will shed some light on them; and perhaps you can edit your post to clarify your question.

The first isomorphism theorem is basically a tautology : it tells you that if you have a surjective morphism and declare "$$x=y$$" precisely when $$f(x)=f(y)$$ then you get an induced map on the new structure when your declaration is true, and that this induced map is injective, and has the same image as the original one. The fact that it has the same image is obvious because there is a factorisation, so I won't mention it. The fact that you get an induced map is also obvious, because if you don't know which antecedent to choose, it doesn't matter, as they all have the same images; so just choose any antecedent.

Finally, the fact that the induced map is injective is also obvious because you've forced it to be ! If $$x,y$$ have the same image in the new structure, then any antecedent of them do too, so they have been declared to be equal ! Therefore $$x=y$$ by the pure will of you, the new structure creator. In other words, the first isomorphism theorem is you wanting a map to be injective, and declaring "it is", and by doing so you simply create a new structure (the quotient structure), on which it is, precisely because you declared it to be.