$\DeclareMathOperator*{colim}{colim}$ $\DeclareMathOperator{Hom}{Hom}$
Let $X$ be a (compactly generated) space. Let $\mathcal{C}=\{D\subset X:D\text{ is compact}\}$
Let $A$ be a compact Hausdorff space. (In my question, $A$ is $|\Delta^n|$, the standard topological $n$-simplex)
Since continuous map sends compact space to compact space, we have an bijection in Set:
$\colim_{D\in\mathcal{C}}\mathrm{Hom}(A,D)\cong\mathrm{Hom}(A,X)$
Since $A,D,X$ are compactly generated spaces, the Hom sets can be equipped with compact-open topology.
I want to know whether the bijection above is a homeomorphism in Top.
The map $\colim_{D\in\mathcal{C}}\mathrm{Hom}(A,D)\to\mathrm{Hom}(A,X)$ induced by UMP is continuous and bijection.
It remains to show that whether it's an open map.
Let $W\subset\colim_{D\in\mathcal{C}}\mathrm{Hom}(A,D)$ be an open subset. This is equivalent to that for each $D\in\mathcal{C}$, $W\cap\Hom(A,D)$ is open.
For $f_0\in W$, we shall find an open neighborhood of $f_0$ in $\Hom(A,X)$
For each $D$, there is compact $K_D\subset A$, open $U_D\subset D$, s.t. $V(K_D,U_D)=\{f:f(K_D)\subset U_D\}\subset W$
Each $U_D$ needn't be open in $X$, and it seems that $V(K_D,U_D)$ needn't be open in $\mathrm{Hom}(A,X)$
In my intuition, if $U_D$ shrinks as $D$ expanding, then $f_0$ couldn't has an open nbhd in $\mathrm{Hom}(A,X)$. But could this happen?
$\DeclareMathOperator*{\colim}{colim}$ $\DeclareMathOperator*{\Hom}{Hom}$ I found a proof in the paper "The Category of CGWH Spaces" by N.P.Strickland
A filtered diagram $\{A_i\}$ of closed inclusions is called strongly filtered, if every compact subset of $\colim{A_i}$ is contained in some $A_i$.
We shall prove the following fact: If $A$ is compact, then $\Hom(A,-)$ preserves strongly filtered colimits.
i.e. for all strongly filtered diagram $\{X_i\}$, $\colim\Hom(A,X_i)\cong\Hom(A,X)$ is homeomorphism.
The proof is based on the Yoneda lemma, by the observations in my question, the natural map $\colim\Hom(A,X_i)\to\Hom(A,X)$ is a continuous bijection.
For each compact $W$, we have bijection: $\Hom(W,\colim\Hom(A,X_i)\cong\Hom(W,\Hom(A,X))$ in Set.
This is obtained by an easy diagram chasing.
Use the commutativity between $\Hom(-,X_i)$ and filtered colimits in Set, this bijection holds for even noncompact spaces.
Now by Yoneda's Lemma, the proof is completed.