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Let $X$ and $Y$ be CW complexes. I wish to consider the mapping space $\mathbf{map}(X,Y)$ of continuous maps $X\to Y$, equipped with the (compactly generated version of the) compact-open topology. I believe the path-components and the connected components of this space coincide. (This would be true if $\mathbf{map}(X,Y)$ were a delta-generated or numerically-generated topological space, but I’ve been unable to verify this.) (This came up from reading Lurie’s definition of the homotopy category of spaces in HTT; in order for his definition to be the right one, the path components of $\mathbf{map}(X,Y)$ must coincide with the connected components.)

Tomo
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1 Answers1

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No, not in general. For instance, consider what happens if $Y=\{0,1\}$ and $X$ is an infinite discrete space. Then the mapping space is just $Y^X$ with the product topology, which is not locally connected and therefore not delta-generated. Or, consider if $Y=[0,1]$ and $X$ is an uncountable discrete space. Then again, the mapping space is $Y^X$ with the product topology, which is not sequential and therefore not delta-generated.

Less trivially, there are in fact examples where the path-components and connected components of $\mathbf{map}(X,Y)$ do not coincide. In particular, consider any example where $X$ is a countable CW complex, $Y$ is connected, and there exists a phantom map $f:X\to Y$, a map whose restriction to each finite subcomplex of $X$ is nullhomotopic but such that $f$ is not nullhomotopic. Write $X$ as an increasing union of finite subcomplexes $K_n$. For each $n$, we can find a map $g_n:X\to Y$ which coincides with $f$ on $K_n$ but which is nullhomotopic. These maps $g_n$ then form a sequence which converges to $f$ in the compact-open topology, and thus also in its $k$-ification (since a convergent sequence together with its limit forms a compact set). So, $f$ is in the closure of the path-component of $\mathbf{map}(X,Y)$ consisting of nullhomotopic maps, but is not in that path-component, so that path-component is not a component.

It is true that $\mathbf{map}(X,Y)$ is delta-generated if $X$ is a finite CW complex. In that case, note first that $\mathbf{map}(X,Y)$ is the colimit of $\mathbf{map}(X,K)$ where $K$ ranges over finite subcomplexes of $Y$ (since every compact subset of $Y$ is contained in a finite subcomplex), so it suffices to consider the case where $Y$ is also finite. Then $Y$ is a retract of an open subset $U\subseteq\mathbb{R}^n$ for some $n$, and the topology of $\mathbf{map}(X,Y)$ is just the topology of uniform convergence with respect to the metric on $Y$ induced from the Euclidean metric. Now for any $f:X\to Y$ and any $g:X\to Y$ sufficiently close to $f$, the linear homotopy from $f$ to $g$ is contained in $U$. Moreover, if $g$ is sufficiently close to $f$, every stage of the composition of this linear homotopy with the retraction $U\to Y$ is close to $f$ (since the retraction is uniformly continuous on any compact set). That is, for any neighborhood $V\subseteq \mathbf{map}(X,Y)$ of $f$, there is a neighborhood $W\subseteq V$ of $f$ such that every element of $W$ is connected to $f$ by a path in $V$. This condition together with first-countability of $\mathbf{map}(X,Y)$ implies it is delta-generated (by a minor modification of the argument for $1\Rightarrow 2$ here).

Eric Wofsey
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  • Hm, thanks. I still don’t know, then, how to see that Lurie’s Example 1.1.3.3 in HTT is the right definition of the homotopy category of spaces. – Tomo Nov 20 '21 at 05:12
  • @Tomo: What do you think is wrong with it? – Eric Wofsey Nov 20 '21 at 05:39
  • He defines the morphisms to be $\pi_0$ of the mapping space. But the quotient by homotopy equivalence should identify only up to path component. My friend says that one should just interpret the mapping space object as the corresponding one in the category of numerical generated spaces (applying the coreflector). If you do this then it is fine, but he doesn’t mention numerically generated spaces. – Tomo Nov 20 '21 at 05:47
  • $\pi_0$ is the set of path-components, not the set of components. – Eric Wofsey Nov 20 '21 at 05:49
  • Ah ok, well that makes sense then! – Tomo Nov 20 '21 at 05:55