Abstract nonsense proof

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\eta\cdot\gamma$.)

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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 $G$).

Answer 4

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.