Do Homotopy Groups commute with generalized filtered colimits? 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 $F(i)$. 

The answer is no, not even $\pi_1$ does:
If $X$ is a metric space and $Y$ any topological space, then a map $f\colon X\to Y$ is continuous if and only if for every convergent sequence $(x_n)_{n\in\mathbb{N}}$ in $X$, the sequence $(f(x_n))_{n\in\mathbb{N}}$ converges with limit point $f(\lim_n x_n)$.
Furthermore, if $X$ is compact, the closure of the set underlying a convergent sequence is closed and countable.
Thus, such an $X$ is naturally isomorphic to the filtered colimit of all of its countable, closed subsets.
Fixing a point $x_0\in X$, we may equivalently restrict the indexing category to the set of countable, closed subsets containing $x_0$.
Now take a space as easy as $X = S^1$ with any point $x_0\in X$.
We have $\pi_1(A,x_0) = 1$ for each closed and countable subset $A\subset X$, but $\pi_1(X,x_0) = \mathbb{Z}$, of course.
