TITLE: On limits and Colimits
QUESTION [8 upvotes]: I want to ask a stupid question. Let $I$ be an infinite set and suppose $i$ belongs to $I$. I wonder whether following morphisms exist in general:
Hom($A$,colim $B_i) \to$ lim Hom($A,B_i$) and 
Colim Hom($A,B_i) \to$ Hom($A$,colim $B_i$)
What I know is: if we replace lim by infinite product and colim by infinite coproduct, it exists. But I am not sure in this general case above.

REPLY [21 votes]: For any diagram $B_i$ and an object $A$ in a category, there are natural maps of sets:

colim Hom($A,B_i) \to$ Hom($A$, colim $B_i$)
colim Hom($B_i,A) \to$ Hom(lim $B_i, A$)

These maps need not be isomorphisms, in general (neither even when the diagram is filtered, nor when it is finite).  Nor are they isomorphisms for infinite products and coproducts, in general (for finite products and coproducts in an additive category they are isomorphisms, though).
Besides, for any diagram $B_i$ and an object $A$ there are natural isomorphisms of sets:

Hom($A$, lim $B_i$) = lim Hom($A,B_i$)
Hom(colim $B_i, A$) = lim Hom($B_i,A$)

These isomorphisms hold for any diagram (it does not have to be filtered, nor does it have to be finite).  Actually, they hold by the definition of lim and colim.

REPLY [3 votes]: In general, your category has to admit small limits for that to even start to begin to make any sense at all.  Also, as I noted in my question, I'm fairly sure that you've got it backwards.  It should be:
Hom(colim(F(-)),X) is isomorpic to lim Hom(F(-),X), and Hom(Y,lim(F(-))) is isomorphic to lim Hom(Y,F(-)), where we're limiting and colimiting over the domain of F, where F is a functor into our category from some other category (Diagrams for example.)
I don't know if this is what you actually wanted, but if I read your question the way you typed it out, the first one doesn't make sense, since covariant hom is covariant.  The second one might be true provided that the limit has certain restrictions on it or if covariant hom has an appropriate adjoint.  There might be other cases, but it's not true in general.  If you're just looking for the existence of a map in the second one, then it's trivial.