I know that if X is a topological space such that $X= \underset{i}{\bigcup} X_i$ where $X_0 \subset X_1 \subset ... \subset X_n \subset ...$, where $X_i$ are all hausdorff, then the functor $\pi_n(\_)$ commutes with the colimit: $$\varinjlim \pi_n(X_i, x_0)\cong \pi_n(X, x_0) $$ One way to prove this is to show that each continuous map from a compact space K into X factors over some $X_i$.

Can this be generalized to all filtered colimits? Thus, if $I$ is a filtered category, F a functor $I \longrightarrow$ **Top**, does it generally hold that: $$\varinjlim \pi_n(F(i), x_0)\cong \pi_n(\varinjlim F, x_0) $$ If yes, how can I prove this? I have tried to generalize the proof of the above statement, but I do not know how to prove that each continuous map from a compact space K into the limit factors over some $F(i)$.