Show that $\lim$ is a right adjoint to the constant functor

A problen in category theory:

Let $$I$$ be a small category and $$\mathcal C$$ be a comolete category, and consider the functor $$\lim : Func(I, \mathcal C) \to \mathcal C$$. Show that $$\lim$$ is right adjoint to the functor $$G: \mathcal C \to Func(I, \mathcal C)$$ that takes an object to the constant functor and a morphism to the corresponding natural transformation between constant functors.

While this looked like a simple exercise, I am stuck on it.

Let $$X, Y \in \mathcal C$$, and $$F$$ a functor from $$I$$ to $$\mathcal C$$.

We have to show that there is some isomorphism $$\phi : Hom(F_X, F) \cong Hom(X, \lim F)$$ where $$F$$ is a functor from $$I$$ to $$\mathcal C$$ and the image of $$F_X$$ consists of only the element $$X$$ and the identity morphism. Of course, by definition of $$\lim F$$ there is one and only one morphism from $$X$$ to $$\lim F$$, so $$\phi$$ is directly determined. However, I am confused as to why $$\phi$$ has to be an isomorphism and natural in both components. Indeed, there can be a lot of natural transformation from $$F_X$$ to $$F$$... am I missing something? I showed that such a $$\phi$$ must saitsfy $$\phi(\mathcal E \circ \eta^f) = \phi(\mathcal E) \circ f$$ where $$\mathcal E$$ is a natural transformation from $$F_Y$$ to $$F$$.

It seems like I have to somehow show that there is only one possible natural transformation from $$F_X$$ to $$F$$...

• There is not, generally, only one morphism $X\to\lim F$; if there were you'd be correct that you'd need to show there was exactly one natural transformation $F_X\to F$. There is exactly one $X\to\lim F$ per commuting cone under $X$. – Malice Vidrine Oct 21 '18 at 23:28

So to use $$G$$ in the sense of the quoted text, $$\phi:\hom(GX,F)\to \hom(X,\lim F)$$ just takes a cone $$\alpha:GX\to F$$ to the unique $$\bar{\alpha}:X\to \lim F$$ required by the universal property. The inverse, $$\phi^{-1}:\hom(X,\lim F)\to \hom(GX,F)$$, takes an $$f:X\to\lim F$$ to the composition $$\pi\circ Gf:GX\to F$$, where $$\pi$$ is the limiting cone $$G(\lim F)\to F$$. That is, it takes $$f$$ to the cone whose component at $$A\in \mathrm{Obj}(I)$$ is $$\pi_A\circ f$$. That these are inverses to each other follows readily from the existence and uniqueness conditions of the universal property.
The reason you want this to be an isomorphism is that given any cone $$\gamma:GA\to F$$, a morphism $$X\to A$$ results in a new cone $$GX\to F$$ by composition. But all this gives you is a function $$\hom(X,A)\to\hom(GX,F)$$. But if $$A$$ is a limit, the universal property says that this function is injective (no two morphisms $$X\to A$$ give you the same cone), and reversible (every cone $$GX\to F$$ is a composition of a morphism $$X\to A$$ with the limit projections).
And you want it to be natural because otherwise you would have a situation where (among other things), for $$f:X\to\lim F$$ and $$g:Y\to X$$, the cone coming from the composition $$\pi\circ Gf\circ Gg:GY\to F$$ need not be the same as the cone corresponding to $$f\circ g:Y\to \lim F$$, and that's definitely not what we want.