Category Theory: homset preserves limits 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?
 A: There is a rather silk proof which requires some observations:


*

*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 $F$.

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

*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. 
A: page 9 of http://www.math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Henderson.pdf
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.
