Category Theory – Homset Preserves Limits

category-theorylimits-colimitsrepresentable-functor

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?

Best Answer

There is a rather silk proof which requires some observations:

  1. Let $F$ be a diagram in $\mathcal{C}$ indexed by $\mathscr{J}$. Then, a limiting cone $(\lim F, \mu\colon \Delta \lim F \to F)$ corresponds exactly to an isomorphism $$ \mathcal{C}(X, \lim F) \cong [\mathscr{J}, \mathcal{C}](\Delta X, F). $$ natural in $X$ where the right hand side is just the set of cones over $F$.

  2. The set of $F$-cones with tip $B$ is isomorphic to $[\mathcal{J}, \mathbf{Set}](\Delta 1, \mathcal{C}(B, F-))$.

  3. When $\mathcal{C} = \mathbf{Set}$ and $X = 1$ in the first isomorphism, we have $$ \lim F \cong [\mathcal{J}, \mathbf{Set}](\Delta 1, F) $$ i.e. the set of cones from a singleton set $1$ over $F$ is the limit of $F$.

Finally, we arrive the conclusion that the set $\mathcal{C}(B, \lim F)$ is the limit of $\mathcal{C}(B, -) \circ F$ from $$ \mathcal{C}(B, \lim F) \cong [\mathcal{J}, \mathcal{C}](\Delta B, F) \cong [\mathcal{J}, \mathbf{Set}](\Delta 1, \mathcal{C}(B, F-)) \cong \lim \mathcal{C}(B, F-) $$ where the first isomorphism follows from our first observation, the second from the second observation, and the third from the inverse of the third observation.

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