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Does the functor $S:\mathbf{Top}\to \mathbf{sSets}$ given by $S(X)_m=Hom_{\mathbf{Top}}(\Delta^m,X)$ preserve colimits? If not, what is a counterexample?

The only things I can say are that it preserves limits because it is a right adjoint and that it preserves filtered colimits since each $\Delta^m$ is a finite CW-complex.

Does the functor $S$ preserve homotopy colimits? If not, what is a counterexample?

Again I can only say that it preserves homotopy limits as because it is a right Quillen adjoint. This Quillen adjunction with left adjoint the realization functor however is also a Quillen equivalence. I wonder if it follows already from this fact, that the right adjoint preserves homotopy colimits (and that the left adjoint preserves homotopy limits).

Ronald Bernard
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2 Answers2

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I find the claim about filtered colimits suspect – that question is very subtle (which is why the small object argument in $\mathbf{Top}$ has to be done carefully). Anyway, $S$ preserves coproducts, because the spaces $\Delta^n$ are all connected. Thus, $S$ preserves colimits of diagrams of discrete topological spaces. (In fact, $S$ restricts to an equivalence between discrete topological spaces and discrete simplicial sets.)

However, consider the Sierpiński space $X$, i.e. the space with one open point and one closed point. Take the coequaliser of the two maps $1 \to X$; it is $1$ again, of course. Now, $S(X)_n$ is the set of open subsets of $\Delta^n$, and the coequaliser of the two maps $S(1)_n \to S(X)_n$ simply identifies the element corresponding to $\emptyset$ with the element element corresponding to $\Delta^n$; this, of course, is still an infinite set. So we have the required example of a coequaliser not preserved by $S$.

Note, $X$ is compactly generated, so not completely pathological! But if you prefer to work in the category of compactly-generated Hausdorff spaces (say), here is another counterexample. Consider two maps $1 \to \Delta^1 \amalg \Delta^1$ whose coequaliser is $\Delta^1$. Suppose, for a contradiction, that $\mathbf{Top}(\Delta^1, -)$ preserves this coequaliser. Then the induced map $\mathbf{Top}(\Delta^1, \Delta^1 \amalg \Delta^1) \to \mathbf{Top}(\Delta^1, \Delta^1)$ must be surjective, but that is false: since $\Delta^1$ is connected, its image in $\Delta^1 \amalg \Delta^1$ must be entirely contained in one of the two copies, so there is no map $\Delta^1 \to \Delta^1 \amalg \Delta^1$ whose composite with the coequaliser map $\Delta^1 \amalg \Delta^1$ is the identity.

As for homotopy colimits: since $S$ is a homotopical equivalence (i.e. preserves weak equivalences and induces an equivalence of homotopy categories), it indeed preserves all homotopy colimits. Here I mean homotopy colimits in the sense of left derived functors.

Zhen Lin
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The first question: This is usually not the case. For example, you can find a counterexample from that the Yoneda functor does not preserve colimits.

Since the colimit in $sSet$ is computed pointwise, one looks for $X_1, X_2, X_0, X_3=X_1\cup_{X_0} X_2$ such that $\hom(\Delta^1, X_i)$ is not a colimit diagram. Let $$X_1=[0,1],\quad X_2=[-1,0],\quad X_0=0,\quad X_3=[-1, 1].$$ Then it is clear that $X_3$ has more paths than paths of $X_1$ plus paths of $X_2$.

Ma Ming
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