Direct limits in homotopy category

Question

It is known that the homotopy category $\mathrm{HoTop}$ is not complete nor cocomplete. Moreover, it also fails to have filtered colimits in general. I am wondering if the universal property of the following "nice" colimit is true:

Let $X_0\to X_1\to X_2\to\cdots$ be a diagram of topological spaces with arrows being inclusions. Let $\varinjlim X_n$ be the direct limit of this diagram in $\mathrm{Top}$. For any space $Y$, suppose we have homotopy equivalent maps $f_n\simeq g_n:X_n\to Y$ for all $n$, and collections of maps $\{f_n\}_n,\{g_n\}_n$ both fit in the diagram. Is it necessarily true that their corresponding induced maps $f,g:\varinjlim X_n\to Y$ are homotopy equivalent?

We can further assume that all the spaces are CW complexes. If in general it is also false, then is there any further assumptions we may need for it being true? Any comments would be very much appreciated.

Answer 1

This is the classical $\mathrm{lim}^1$ phenomenon: While $\mathrm{Map}(X,Y) = \operatorname{lim}\mathrm{Map}(X_n,Y)$ (a homotopy limit), on homotopy classes it is not true that $[X,Y] \cong \operatorname{lim}[X_n,Y]$. Instead, the map $[X,Y]\to \operatorname{lim}[X_n,Y]$ is surjective, and under suitable assumptions has kernel described by $\operatorname{lim}^1 [X_n,\Omega Y]$.

For an explicit example, let $X_n = S^2$, and $X_n\to X_{n+1}$ the degree $n$ map $S^2\to S^2$. Then the homotopy colimit (or, if you prefer, replace $X_n$ by the $n$-th stage of some telescope and take the actual colimit) is a Moore space $M(\mathbb{Q},2)$. By the universal coefficient theorem, $H^3(X;\mathbb{Z}) = \mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})$ is nontrivial. A nontrivial element here corresponds to some nontrivial map $X\to K(\mathbb{Z},3)$. But of course, any two maps $X_n\to K(\mathbb{Z},3)$ are homotopic, since $X_n\simeq S^2$.

Via the $\mathrm{lim}^1$-sequence, this comes from the fact that $\mathrm{lim}^1(\ldots \xrightarrow{n} \mathbb{Z}\xrightarrow{n-1}\ldots)$ is nontrivial (in fact, agrees with $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})$).