# Abstract nonsense proof

What is a simple example of an "abstract nonsense" proof in category theory. For a theorem you are proving, it doesn't matter if the category or regular proof came first, it is just that the category one should be simpler or more elegant. Note that it doesn't matter if the non-category prerequisites are some what complicated. I am sort of looking of how category theory can transfer proofs from one field to another. It should be at a rather simple level if possible.

• Read any of my math.SE posts in abstract algebra ... ;) Jun 7, 2014 at 9:46
• – user55315
Jun 8, 2014 at 14:43
• The shorter part of math.stackexchange.com/a/500193/75923 Jun 14, 2014 at 17:37

One of my favourite :

The fundamental group of a topological group is abelian.

Proof. The fundamental group functor $\pi_1 \colon \mathsf{Top}_\bullet \to \mathsf{Grp}$ from the category of pointed topological spaces to the category of groups preserves products, hence group objects. It then maps group objects of $\mathsf{Top}_\bullet$, i.e. pointed topological groups, to group objects of $\mathsf{Grp}$, i.e. abelian groups (a group is abelian if and only if the inversion is a group morphism). CQFD.

(The usual proof consists of the explicit construction of a homotopy between the concatenation $\gamma \cdot \eta$ of loops $\gamma,\eta$ and the loop $t \mapsto \gamma(t)\eta(t)$, and then another one between the later and $\eta\cdot\gamma$.)

As spotted in the comments, my justification of group objects in $\mathsf{Grp}$ is a bit unclear. Take $G$ a group with group morphisms $m \colon G \times G \to G$, $i \colon G \to G$ and $e \colon 1 \to G$ making it a group object : then $e$ (as a group morphism) select the group unit $1_G$ in $G$ ; hence $$m(g,h) = m((g,1_G)(1_G,h)) = m(g,1_G) m(1_G,h) = gh ;$$ so $i \colon G \to G$ is the true inversion (the one in $G$).

• There is more to "group objects in $\mathsf{Grp}$ are abelian groups" than what you said. In theory there are two group structures on a group object in $\mathsf{Grp}$, and the second inversion is only a morphism wrt the first structure. You really need a variation of the Eckmann-Hilton argument to conclude... Which is exactly the explicit proof you mention later. There's no such thing as free lunch. Jun 8, 2014 at 18:23
• @NajibIdrissi The Eckmann-Hilton argument (at least the part telling the two products are the same) is quite easy, but I'll edit to make things clear. And it is very much not the same as the proof I mention (for which one have to exhibit the homotopy, this is dirty). This might not be a free lunch, but it surely looks like a cheap meal in a fancy restaurant. :)
– Pece
Jun 9, 2014 at 9:09
• It's true that you can somewhat skip a step here, because these are group objects and so you get the inversion etc. But the Eckmann-Hilton argument itself implies that the structure is commutative, and it's derived exactly like the explicit homotopy you speak of. It's also applicable to H-spaces and not just topological groups. Jun 9, 2014 at 9:55

Well, there's Lawvere's rather nice proof of a form of Cantor's theorem; namely, if an object $X$ can index all its own morphisms $X\to A$, then every endomorphism of $A$ has a fixed point.

Assume $e:X\times X\to A$ has the universal property of an evaluation morphism, and let $f$ be any morphism $A\to A$. We form the composite $m:=f\circ e\circ\langle id_X, id_X\rangle:X\to A$. By the universal property for $X$ and $e$, there's a unique morphism $q:1\to X$ with $e\circ \langle id_X, q!\rangle=m$. If we do a little arrow chasing, however, we can see that $m\circ q=f\circ m\circ q$, and $f$ therefore has a fixed point; but $f$ was arbitrary, so any $f$ would have a fixed point.

Two examples are: proof that the free group $FX$ on $X$ is generated by the image of the canonical map $X\rightarrow FX$ and proof that the abelianization of free group is the free abelian group. This way of proving comes up all over the place if one deals properly with the free group, tensor product and even the polynomial ring. Also an argument that is used very very often is: right adjoints preserve limits and left adjoints preserve colimits. In fact the second proof linked above is such an example.

Bezout's identity can be proven by categorical abstract nonsense.

Proposition. For $$m,n\in \mathbb{N}_{\geq 0}$$ we have $$m\mathbb{Z}+n\mathbb{Z}=\operatorname{gcd}(m,n)\mathbb{Z}$$.
Proof. We can consider any poset $$(X,\leq)$$ as a category $$\underline{X}$$ by letting $$\operatorname{Hom}_{\underline{X}}(x,y):= \begin{cases} \{\ast\} & \text{if } x \leq y \\ \ \ \emptyset & \text{else}.\end{cases}$$ The limit (resp. colimit) of a diagram $$D\colon J\rightarrow \underline{X}$$ is the supremum (resp. infimum) of the set $$\{D(j)\}_{j\in J}$$, if it exists. Given another poset $$Y$$, a functor $$\underline{X}\rightarrow \underline{Y}$$ is precisely an order-preserving function $$X\rightarrow Y$$.

Write $$\mathbb{N}= \mathbb{N}_{\geq 0}$$. Consider the divisibility lattice of $$\mathbb{N}$$. The infimum in this poset is given by the greatest common divisor. Let $$\operatorname{Sub}(\mathbb{Z})$$ denote the partially ordered set of $$\mathbb{Z}$$-submodules of $$\mathbb{Z}$$ ordered by inclusion. The supremum in this poset is given by the sum of submodules.

Define the functor $$F\colon \underline{\mathbb{N}}^{opp}\rightarrow \underline{\operatorname{Sub}(\mathbb{Z}})$$ given by $$F(n):=n\mathbb{Z}$$. This is an equivalence of categories since $$\mathbb{Z}$$ is a PID. In particular, it preserves limits and Bezout's identity follows, QED.