Let $C$ and $J$ be categories. Suppose that the limit functor, $\varprojlim$ say, of $C^{J}$ exists, i.e. the limit of any functor $J \to C$ exists.
I would like to prove that $\varprojlim$ is right adjoint to the diagonal functor $\Delta \colon C \to C^J$ by defining a natural isomorphism
$$\Phi \colon \text{Hom}_C(-, \varprojlim-) \to \text{Hom}_{C^J}(\Delta-,-).$$
$\textbf{My idea:}$ For $Y$ in $C$ and $X$ in $C^J$ define the map of sets
$$ \Phi_{Y,X} \colon \text{Hom}_C(Y, \varprojlim X) \to \text{Hom}_{C^J}(\Delta Y,X)$$ by sending a morphism $\alpha \colon Y \to \varprojlim X$ in $C$ to the natural transformation $\beta \colon \Delta Y \to \Delta \varprojlim X$ given by $\beta_j = \alpha$ for all $j$ in $J$.
This map is clearly injective, but how do I see that it is surjective? Surjectivity would mean that any natural transformation $(\beta_j \colon Y \to X(j))_j$ is of the form $(Y \to \varprojlim \tilde{X})_j$, seems quite restricting.
Best Answer
Let $F\in C^J$ and denote by $X = \varprojlim F\in C$. This is the same thing as saying that for every element in the image of $F$ in $C$ we have a morphism to $X$ making the obvious diagram commute and satisfying the usual universal property, which again is the same thing as a morphism $\phi:\Delta X\to F$ in $C^J$ such that every morphism $\Delta Y\to F$ splits through $\phi$.
Now let $\psi\in\hom_C(Y,X)$. Then we naturally have $\Delta\psi\circ\phi:\Delta Y\to F$. The converse is given by the splitting property (which is equivalent to the universal property of limits).