# The Unapologetic Mathematician

## Mathematics for the interested outsider

We can easily see that limits commute with each other, as do colimits. If we have a functor $F:\mathcal{J}_1\times\mathcal{J}_2\rightarrow\mathcal{C}$, then we can take the limit $\varprojlim_{\mathcal{J}_1\times\mathcal{J}_2}F$ either all at once, or one variable at a time: $\varprojlim_{\mathcal{J}_1}\varprojlim_{\mathcal{J}_2}F=\varprojlim_{\mathcal{J}_2}\varprojlim_{\mathcal{J}_1}F$. That is, if the category $\mathcal{C}$ has $\mathcal{J}$-limits, then the functor $\varprojlim_{J}$ preserves all other limits.

But now we know that limit functors are right adjoints. And it turns out that any functor which has a left adjoint (and thus is a right adjoint) preserves all limits. Dually, any functor which has a right adjoint (and thus is a left adjoint) preserves all colimits.

First we need to note that we can compose adjunctions. That is, if we have adjunctions $F_1\dashv F_2:\mathcal{C}\rightarrow\mathcal{D}$ and $G_1\dashv G_2:\mathcal{D}\rightarrow\mathcal{E}$ then we can put them together to get an adjunction $G_1\circ F_1\dashv F_2\circ G_2:\mathcal{C}\rightarrow\mathcal{E}$. Indeed, we have
$\hom_\mathcal{E}(G_1(F_1(C)),E)\cong\hom_\mathcal{D}(F_1(C),G_2(E))\cong\hom_\mathcal{C}(C,F_2(G_2(E)))$

We also need to note that adjoints are unique up to natural isomorphism. That is, if $F\dashv G_1:\mathcal{C}\rightarrow\mathcal{D}$ and $F\dashv G_2:\mathcal{C}\rightarrow\mathcal{D}$ then there is a natural isomorphism $G_1\cong G_2$. This is essentially because adjunctions are determined by universal arrows, and universal arrows are unique up to isomorphism.

Okay, now we can get to work. We start with an adjunction $F\dashv G:\mathcal{C}\rightarrow\mathcal{D}$. Given another (small) category $\mathcal{J}$ we can build the functor categories $\mathcal{C}^\mathcal{J}$ and $\mathcal{D}^\mathcal{J}$. It turns out we get an adjunction here too. Define $F^\mathcal{J}(S)=F\circ S$ for each functor $S:\mathcal{J}\rightarrow\mathcal{C}$. The unit $\eta:1_\mathcal{C}\rightarrow G\circ F$ induces a unit $\eta^\mathcal{J}_S=\eta\circ1_S:S\rightarrow G\circ F\circ S$. We can similarly define $G^\mathcal{J}$ and $\epsilon^\mathcal{J}$, and show that they determine an adjunction $F^\mathcal{J}\dashv G^\mathcal{J}:\mathcal{C}^\mathcal{J}\rightarrow\mathcal{D}^\mathcal{J}$

Now let’s say that $\mathcal{C}$ and $\mathcal{D}$ both have $\mathcal{J}$-limits. Then we have an adjunction $\Delta\dashv\varprojlim_\mathcal{J}:\mathcal{C}\rightarrow\mathcal{C}^\mathcal{J}$ and a similar one for $\mathcal{D}$. We can thus form the composite adjunctions
$F^\mathcal{J}\circ\Delta\dashv\varprojlim_\mathcal{J}\circ G^\mathcal{J}:\mathcal{C}\rightarrow\mathcal{D}^\mathcal{J}$
$\Delta\circ F\dashv G\circ\varprojlim_\mathcal{J}:\mathcal{C}\rightarrow\mathcal{D}^\mathcal{J}$

So what is $F^\mathcal{J}(\Delta(C))$? Well, $\Delta(C)$ is the functor that sends every object of $\mathcal{J}$ to $C$ and every morphism to $1_C$. Then composing this with $F$ gives the functor that sends every object of $\mathcal{J}$ to $F(C)$ and every morphism to $1_{F(C)}$. That is, we get $\Delta(F(C))$. So $F^\mathcal{J}\circ\Delta=\Delta\circ F$. But these are the two left adjoints listed above. Thus the two right adjoints listed above are both right adjoint to the same functor, and therefore must be naturally isomorphic! We have $\varprojlim_\mathcal{J}G\circ T\cong G(\varprojlim_\mathcal{J}T)$ for every functor $T:\mathcal{J}\rightarrow\mathcal{D}$. And thus $G$ preserves $\mathcal{J}$-limits.