Let $F:J \rightarrow \mathrm{Top}$ be a diagram in the category of topological spaces and $X:=\mathrm{colim} F$ be its colimit. For each $a \in \mathrm{ob}(J)$, denote by $\imath_a:F(a)\rightarrow X$ the resulting inclusions. Let $I$ be a full subcategory of $J$. Define a topological subspace $A\subseteq X$ via $A := \bigcup_{a \in \mathrm{ob}(I)} \imath_a(F(a))$. The question is: is $A$ equal to $\mathrm{colim} F|_I$?
I can only find an epimorphism $\mathrm{colim} F|_I \rightarrow A$. This map comes from the universal property of the colimit, and it is an epimorphism due to the fact that there are epimorphisms $\bigsqcup_{a \in \mathrm{ob}(I)}F(a) \rightarrow \mathrm{colim} F|_I$ and $\bigsqcup_{a \in \mathrm{ob}(I)}F(a) \rightarrow A$. Is there a simple categorical way of showing that the map is an isomorphism?
I could probably use the explicit description of the colimits, but I'm trying to find an aesthetically pleasing proof (if the statement is true at all). One could also pose this question in general, in the setting of an arbitrary cocomplete category $\mathcal{C}$: if there is a diagram $F:J \rightarrow \mathcal{C}$ and a full subcategory $I \subseteq J$, can one find a monomorphism $\mathrm{colim} F|_I \rightarrow \mathrm{colim}F$?
This is false in general. For a counterexample, consider the pushout square $$ \require{AMScd} \begin{CD} S^1 @>>> D^2\\ @VVV @VVV\\ * @>>> S^2 \end{CD} $$ in $\mathsf{Top}$. We let $J$ be the category $1\leftarrow 0\rightarrow 2$ and let $F$ be the functor sending $J$ to the diagram $*\leftarrow S^1\rightarrow D^2$. Let $I$ be the full subcategory on the object $0$. Then the colimit of $F|_I$ is just $S^1$, while the image of $S^1$ in $S^2$ in the pushout square above is a single point.