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edit I updated my question at the end, I think the claim may be false?

Let $(L,\lambda)$ be a limit cone of a diagram $D$ in a category.

For any object $X$ it is said that the hom functor $Hom(X,-)$ preserves limits.

How can I prove $(Hom(X,L),\lambda \circ -)$ is a limit cone in Set?

Because Set has all limits, I tried to build the limit there and got a universal map from it to that cone, but I need to show that map is an isomorphism to say its the limit cone.

I think this might not work at all, what is a nice way to prove it?

I got further with a simple example, products and found this:

  • $Hom(X,A\times B) \to Hom(X,A)\times Hom(X,B)$ by universal map
  • $Hom(X,A)\times Hom(X,B) \to Hom(X,A\times B)$ let $(f,g) \in Hom(X,A)\times Hom(X,B)$ then $x \mapsto (fx,gx) \in Hom(X,A\times B)$ and the legs of the cones commute because $f = x \mapsto f x$.

The composition of both maps $Hom(X,A)\times Hom(X,B) \to > Hom(X,A)\times Hom(X,B)$ is the identity because every self map from a limit that makes legs commute is the identity.

But how do I show the composition is the identity other way around? Is $Hom(X,A\times B) \to Hom(X,A\times B)$ true?

Hopefully this will generalize too.

Thanks to Hurkyl,

If $T$ is a terminal object, then $Hom(X,T)$ is a one element set so it's a terminal object in Set.

Using this idea I also proved the claim for equalizers in the category of finite sets.

If E is the equalizer of finite sets A and B, then $Fin(X,E)$ has $|E|^{|X|}$ elements. If F is the equalizer of $Fin(X,A) \to \to > Fin(X,B)$, but all the $Fin(X,B)$ maps factor through $A$ so that $|F| = > |E|^{|X|}$, then the sets are isomorphic.

Proving it for equalizers would give the theorem for all finite limits by the fact a finite limit can be constructed from these three primitives, but I would like to know:

Is there a uniform proof for an arbitrary diagram?

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Are you asking questions one-by-one from an exercise sheet or something? I'll give you a hint anyway: show that the limit of $\textrm{Hom}(X, D)$ is precisely the set of cones from $X$ to $D$. – Zhen Lin Jan 18 '13 at 21:06
I'm just trying to close the gaps in my understanding. I don't understand your hint because the limit is a single cone not a set of cones. – user58512 Jan 18 '13 at 21:15
this question is because "representable functors preserve limits since they are isomorphic to homset functors, which preserve limits".. but I didn't have proof of that. – user58512 Jan 18 '13 at 21:21
I found a proof here but don't understand it – user58512 Jan 18 '13 at 21:52
If $T$ is terminal, then $\hom(X,T)$ is a set with one element. – Hurkyl Jan 19 '13 at 11:59
up vote 1 down vote accepted

page 9 of

Let $D : \mathcal J \to \mathcal C$ be a diagram for some locally small category $\mathcal C$.

Let $(L,\forall i \in \mathcal J, L \overset{\lambda_i}{\to} D(i))$ be a limit cone of that diagram.

Let $X$ be any object of $\mathcal C$.

Let $(S,\forall i \in \mathcal J, S \overset{f_i}{\to} \mathcal C(X,D(i)))$ be any cone over the diagram $\mathcal C(X,D-) : \mathcal J \to \mathbf{Set}$.

We want to construct a unique map $u : S \to \mathcal C(X,L)$ that makes the legs commute.

Given $s \in S$, we have a cone $(X,\forall i \in \mathcal J, X \overset{f_i(s)}{\to} D(i))$ in $\mathcal C$ and thus a universal map $u_s : X \to L$ such that $f_i(s) = \lambda_i \circ u_s$ for each leg, and the triangles of the cone commute because $f_i(s)$ came from a cone.

This gives us a map in set $u : S \to \mathcal C(X,L)$ that makes all the legs of the cone commute. It is also the unique such map because it is pointwise unique.

That proves the theorem.

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