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The homotopy category $\mathsf{hTop}$ (topological spaces localized at the weak homotopy equivalences) doesn't have many limits / colimits, but it does have products, computed on the point-set level.

Now, on the point-set level, (compactly generated) spaces are cartesian closed. Cartesian closure in some sense only involves finite products, which are preserved in the passage to the homotopy category. So we might hope that the homotopy category has exponentials, computed on the point-set level.

However, there seems to be a problem. Consider the unit interval $I$. Then $I \times (-): \mathsf{hTop} \to \mathsf{hTop}$ is isomorphic to the identity functor. But the exponential functor $(-)^I: \mathsf{hTop} \to \mathsf{hTop}$ sends $X$ to $X^I$, which is isomorphic in $\mathsf{hTop}$ to the discrete space on the path components of $X$, since the path space of a path-connected space is contractible. These functors are clearly not adjoint -- the adjoint of the identity functor is also the identity functor. So if there is an exponential on $\mathsf{hTop}$, it is not computed at the point-set level.

But perhaps this can be salvaged? Is the homotopy category cartesian closed?

EDIT As discussed in the comments, I'm simply confusing based and unbased path spaces, which are very, very different. So most of what I wrote was wrong: the unbased path space $X^I$ is homotopy equivalent to $X$ via the "constant path map" $X \to X^I$. So $I \times(-)$ and $(-)^I$, viewed as functors on $\mathsf{hTop}$, both become isomorphic to the identity functor and hence remain adjoint. So perhaps $\mathsf{hTop}$ does indeed have a cartesian closed structure computed as at the point-set level.

tcamps
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  • "the path space of a path-connected space is contractible" I very much doubt that's true. Am I wrong? – Olivier Bégassat Sep 27 '15 at 15:42
  • @OlivierBégassat: The path space here is based paths, maps $f: I \to X$ with $f(0) = x_0$. If one doesn't base things then $X^I$ is homotopy equivalent to $X$. –  Sep 27 '15 at 15:43
  • @MikeMiller ok, but why then does the OP say that the path space is isomorphic to the discrete set of path components? This doesn't seem to gel with a basepoint approach. – Olivier Bégassat Sep 27 '15 at 15:46
  • The actual exponential in $\mathsf{Top}$ is unbased paths. I guess maybe I am just confusing these two spaces. – tcamps Sep 27 '15 at 15:47
  • If you consider unbased paths, then there is a homotopy equivalence $X\simeq X^I$. – Olivier Bégassat Sep 27 '15 at 15:48
  • Of course, you're totally right. So I guess my "obstruction" to cartesian closure doesn't stand, which answers my main confusion. I suppose the question still stands whether the homotopy category is cartesian closed... – tcamps Sep 27 '15 at 15:52
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    The homotopy category is cartesian closed, and the localisation functor preserves finite products and exponentials. This is not hard to show directly, but it is also a special case of a general result about monoidal model categories. – Zhen Lin Sep 27 '15 at 18:49
  • What is the generalized version of the result? I would guess that you have to derive the tensor and hom, right? For instance, I very much doubt that the smash product preserves weak equivalences of pointed spaces. – tcamps Sep 28 '15 at 16:09

1 Answers1

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Yes.

Let $i_0,i_1:*\to I$ be the inclusions of the endpoints of the interval, and denote composition with these two maps by $ev_0,ev_1:Top(A\times I,Z)\to Top(A,Z)$.

The diagram $$\require{AMScd} \begin{CD} Top(A\times I,X\times Y) @>{ev_0}>> Top(A,X\times Y) \\ @V{-\circ id_A\times 1}VV \\ Top(A,X\times Y) \end{CD}$$ is the same as the diagram $$\begin{CD} Top(A\times I,X)\times Top(A\times I,Y) @>{(ev_0,ev_0)}>> Top(A,X)\times Top(A,Y) \\ @V{(ev_1,ev_1)}VV \\ Top(A,X)\times Top(A,Y) \end{CD}$$ The pushout of the first diagram is $Ho(Top)(A,X\times Y)$, and of the second diagram is $Ho(Top)(A,X)\times Ho(Top)(A,Y)$. This shows that $X\times Y$ still has the universal property of the prouct in the homotopy category.

Similarly, $Ho(Top)(A,X^Y)$ is a certain pushout over $Top(A\times I,X^Y)$, which using the cartesian closedness of $Top$ can be rewritten as the pushout yielding $Ho(Top)(A\times Y,X)$. Thus $X^Y$ has the exponential universal property in $Ho(Top)$ as well as in $Top$, and we conclude the homotopy category is cartesian closed.

Kevin Carlson
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  • Huh, I've never seen this argument before. Though, this is for the classical homotopy category. – Zhen Lin Sep 28 '15 at 00:27
  • @ZhenLin Right, good point. I'm a bit uncomfortable with the unpointed weak model structure, but shouldn't everything be bifibrant? – Kevin Carlson Sep 28 '15 at 04:22
  • Nope – that would imply that everything has the homotopy type of a CW complex. – Zhen Lin Sep 28 '15 at 07:14
  • @ZhenLin yeah, of course. – Kevin Carlson Sep 28 '15 at 16:17
  • How does this argument work? Pushouts don't commute with products. When proving this for myself, I used that finite products commute in $\mathsf{Set}$ with a similar diagram, but expressed as a reflexive coequalizer. For infinite products, one can use that products commute with equivalence relations in $\mathsf{Set}$. – tcamps Sep 28 '15 at 16:49
  • @tcamps I just computed this specific pushout. But thanks for the comment, it's worth pointing out how this fails to generalize. I guess to really do the computation in detail you'd have to use that homotopy is an equivalence relation anyway. – Kevin Carlson Sep 28 '15 at 17:06
  • I updated your diagrams to make them more readable, I hope you don't mind. – Najib Idrissi Sep 30 '15 at 07:29
  • @NajibIdrissi not at all, thanks! I didn't realize we could use packages here. – Kevin Carlson Sep 30 '15 at 19:55
  • I hope you don't mind; I put the yes/no answer up front, so that people looking here for reference can quickly get the answer at a glance. –  Nov 11 '15 at 04:39
  • @Hurkyl thanks for the improvement! – Kevin Carlson Nov 11 '15 at 05:26
  • "...using the cartesian closedness of $\mathbf{Top}$"? But $\mathbf{Top}$ is not cartesian closed. https://math.stackexchange.com/questions/123950/is-top-provably-not-cartesian-closed – Quique Ruiz Apr 24 '24 at 04:06
  • @QuiqueRuiz Presumably I was referring to $\mathbf{Top}$ as the category of compactly generated spaces, as Tim did in the OP. – Kevin Carlson Apr 24 '24 at 20:50