# Elegant demonstration with minimum category theory knowledge

I come from computer science background so Category Theory show up naturally inside the Category Hask (Types and Functions). I studied a little bit of the theory, but I wish to have some examples of mathematical proofs that use it. Do you have examples of (useful/practical) theorem that can proven using Category theory and yet is accessible to someone with little knowleadge of theory (functors, limits, colimits, maybe natural transformations, ...)

I think this will help me understand deeper the usefulness and beauty of category theory !

Thanks :)

• Perhaps this question would help you : math.stackexchange.com/questions/1208375/… – Arnaud D. Jan 3 '18 at 10:00
• You should read about Yoneda's lemma as a generalization of Cayley's theorem in group theory (and therefore a source of proof) – leibnewtz Jan 3 '18 at 10:07

A typical application is of the following kind. At some point in a proof you might need to know whether or not some functor $F$ commutes with limits or colimits of a particular shape. The easiest way to resolve such questions is that

• if $F$ is a left adjoint then it commutes with all colimits, and
• if $F$ is a right adjoint then it commutes with all limits.

Otherwise things are hard. But this is already a very useful observation: for example, it implies that tensor product $M \otimes (-)$ (say, of modules) commutes with colimits, since it is left adjoint to hom $\text{Hom}(M, -)$.

Here is an example; there are many more like it. The symmetric algebra functor $S(V) = \bigoplus_n S^n(V)$ is the left adjoint to the forgetful functor from commutative $k$-algebras to $k$-vector spaces. It follows that it commutes with colimits, and in particular that it sends direct sums of vector spaces to tensor products of $k$-algebras, so we get that

$$S(V \oplus W) \cong S(V) \otimes S(W)$$

(a kind of categorified exponential property). Keeping track of gradings gives the following identity involving symmetric powers:

$$S^n(V \oplus W) \cong \bigoplus_{i+j=n} S^i(V) \otimes S^j(W).$$

An analogous fact is true for exterior powers but it requires slightly more work to state the appropriate universal property.

• For exterior powers, you should consider the category of graded commutative $R$-algebras (GCA). Then the exterior algebra is left adjoint to the forgetful functor $R\text{-GCA} \to R\text{-Mod}$. The coproduct in $R\text{-GCA}$ is given by $(A\otimes B)_n = \bigotimes_{p+q=n} A_p\otimes B_q$ with $(x\otimes y)\cdot (x'\otimes y') = (-1)^{\deg x'\cdot \deg y}\,xx'\otimes yy'$. So $\Lambda (M\oplus N) \cong \Lambda (M)\otimes \Lambda (N)$. – AAA Jan 4 '18 at 5:04
• * By a graded commutative algebra, I mean an $R$-algebra where the multiplication of homogeneous elements satisfies 1) $x\cdot y = (-1)^{\deg x\,\deg y}\,y\cdot x$ and 2) $x\cdot x = 0$ if $x$ is homogeneous of odd degree. – AAA Jan 4 '18 at 5:09
• ...In particular, we get by abstract nonsense isomorphisms $\Lambda^k (M\oplus N) \cong \bigoplus_{p+q = k} \Lambda^p (M) \otimes \Lambda^q (N)$, and hence Vandermonde's identity ${m+n \choose k} = \sum_{p+q=k} {m\choose p}\,{n\choose q}$. – AAA Jan 4 '18 at 5:17
• @AAA: this is incorrect; that's why I said it require slightly more work to state the appropriate universal property. One way to state the appropriate universal property is that the exterior algebra is left adjoint to the functor from graded commutative $R$-algebras to $R$-modules given by taking the graded component of degree $1$ only. Another is to talk about the forgetful functor from graded commutative $R$-algebras to graded $R$-modules; its left adjoint generalizes both symmetric and exterior algebras. – Qiaochu Yuan Jan 4 '18 at 8:19
• I'm not sure the forgetful functor from graded commutative $R$-algebras to $R$-modules even has a left adjoint. – Qiaochu Yuan Jan 4 '18 at 8:23