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From this post and to (co)completness of the category Top of topological spaces and continuous functions we know that for any diagram $B_i$ and an object $A$ in Top, there are natural maps of sets:$\newcommand{\colim}{\operatorname{colim}}$

  1. $\colim C(A,B_i) \to C(A, \colim B_i)$
  2. $\colim C(B_i,A) \to C(\lim B_i, A)$

For any topological spaces, equip $C(X,Y)$ with the compact-open topology; then $C(X,Y)$ is itself a topological space. When is the map of $1$ (resp. $2$) a surjective(or at-least has dense range) continuous map?

Related: This post on the dual question and this post on naturality of the aforementioned maps.

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    $\begingroup$ I guess it rarely happens, but for example, if the diagram is filtered and A is compact, 1 can be homeomorphism. $\endgroup$
    – user43326
    Commented Jan 14, 2020 at 10:11
  • $\begingroup$ Oh, would you happen to have a reference to this? I'd like to read-up on it. $\endgroup$
    – ABIM
    Commented Jan 14, 2020 at 10:20
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    $\begingroup$ No, I don't, but the idea would be, if A is compact, and we have a filtered colimit, and, let's say, the colimit of "good" injective maps (whatever that means), for any map from A to the colimit, the image has to be contained in finite stage. And, if A is not a compact, you can take $B_i$'s to be compact approximation, in which case one sees clearly that the map can't be surjective. $\endgroup$
    – user43326
    Commented Jan 14, 2020 at 10:47

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