Homotopy groups do not, in general, commute with sequential colimits, even for nice maps between nice spaces.
I just learned this beautiful example from Bill Dwyer.
Take the sequence
$S^1\stackrel{2}{\longrightarrow}S^1\stackrel{3}{\longrightarrow}S^1\stackrel{4}{\longrightarrow}\cdots.$
Here $n$ denotes the $n$th power map on $S^1$. Thinking of $S^1$ as $\mathbb{R}/\mathbb{Z}$, one finds that the colimit of this sequence (in the category of topological spaces) is the quotient group $\mathbb{R}/\mathbb{Q}$. Note that this quotient group, topologized as a quotient space of $\mathbb{R}$ by the relation $x\sim y$ if $x-y\in \mathbb{Q}$, has the indiscrete topology. In particular, the colimit of this sequence is a contractible topological space and has trivial homotopy groups.
On the other hand, the colimit of the corresponding sequence of fundamental groups is the group $\mathbb{Q}$ (checking this is a fun exercise).
(There's something sort of odd here, because one might have guessed that $\mathbb{R}/\mathbb{Q}$ would be a model for $K(\mathbb{Q}, 1)$, since after all $\mathbb{R}$ is a free $\mathbb{Q}$-space. But there are no interesting open sets in the quotient and hence no chance of local triviality.)
