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nlab presents a proof that the category of locally compact Hausdorff spaces does not admit infinite products in general. In particular it shows that there is no infinite product of $\mathbb{R}$, since such a product would be a topological vector space, by the universal property of the projection maps, and it would also be a product in $\mathbb{R}\text{Vect}$. It is known that every locally compact Hausdorff topological vector space is finite dimensional, see here for example.

However since linear maps in $\mathbb{R}\text{Vect}$ are not in general proper, such a proof doesn't work in the case that you restrict to proper continuous maps between locally compact Hausdorff spaces.

I wanted to know if there is still a known counterexample that means that this category will not have products?

One reason that suggests to me they might is that topological spaces with local homeomorphisms (étale maps) has products, and they are very weird; for example $\mathbb{R}\times\mathbb{R}^2=∅$ in that category. It also massively restricts what maps can be onto compact spaces (they must come from a compact space, which is another 'intuition' for why it might have products.

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    $\begingroup$ Page 6 of Tyrone Cutler's note here says the product may fail to exist since the projections can fail to be proper (but also remarks that nevertheless at least $\times$ is part of a non-Cartesian monoidal structure on the category of locally compact Hausdorff spaces with proper maps) $\endgroup$
    – Emily
    Commented Mar 8, 2023 at 5:53
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    $\begingroup$ Just want to note - whereas products in the category of topological spaces and local homeomorphisms are indeed weird (there is no terminal object in fact), pullbacks there are quite nice. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 8, 2023 at 6:08
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    $\begingroup$ @Emily I'm not sure that is quite right, it seems to be suggesting that the product would be the same as that in Top, and of course that product doesn't work, but I'm asking if there is a different topological space which is the product in this category. Note that for example, colimits must be different in this category (i.e. the coproduct of locally compact Hausdorff spaces in Top won't be the coproduct in this category, since the *-compactification must preserve colimits into KHaus, and those are just disjoint unions) $\endgroup$
    – Oddly Asymmetric
    Commented Mar 8, 2023 at 6:33
  • $\begingroup$ @OddlyAsymmetric Hmm, now that you mention it I'm not sure whether Tyrone means that the usual product space isn't the product in [LCHaus w/ proper maps] or if the projections failing to be proper somehow imply that [LCHaus w/ proper maps] doesn't have products. (I feel like it's probably the former, but the note says "products in [LCHaus w/ proper maps] don't exist", so again I'm not really sure...). Maybe you could try asking them if they could elaborate (Or hopefully they'll simply see this question :), I've seen Tyrone here (or on MSE?) before) $\endgroup$
    – Emily
    Commented Mar 8, 2023 at 7:04

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Suppose $X,Y$ are locally compact Hausdorff spaces admitting a product $X\otimes Y$ in the category $\mathcal{H}$ of all such spaces. If one of $X,Y$ is empty, then $X\otimes Y$ exists, so I'll assume that neither is.

Then the points of $X\otimes Y$ are exactly the maps $\ast\rightarrow X\otimes Y$. Since $X\otimes Y$ is Hausdorff, any such map is proper. Because of this, both of the categorical projections $$X\xleftarrow{\pi_X} X\otimes Y\xrightarrow{\pi_Y}Y$$ are surjective. They are continuous and proper by assumption.

Working now in $Top$, the category of all spaces, we have the Tychonoff product $X\times Y$ and an induced map $$\theta:X\otimes Y\rightarrow X\times Y$$ factoring the projections. By the above observation, $\theta$ is surjective.

Lemma: Suppose that $f:A\rightarrow B$ and $g:B\rightarrow C$ are maps of spaces $A,B,C$. If $f$ is surjective and $g\circ f$ is proper, then $g$ is proper. $\quad\blacksquare$

Apply the lemma to the composite $$pr_X\circ \theta=\pi_X$$ to conclude that $pr_X$ is proper. This is only possible if $Y$ is compact. Similarly, $X$ must also be compact.

Of course, if $X,Y$ are compact, then the product $X\otimes Y$ exists in $\mathcal{H}$ and is given by the Tychonoff product $X\times Y$.

If $X,Y$ are locally compact Hausdorff and nonempty, then the product $X\otimes Y$ exists in $\mathcal{H}$ if and only if both $X,Y$ are compact.

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  • $\begingroup$ A point of $X\times Y$ is a pair of maps $X\leftarrow *\rightarrow Y$, and this is exactly a point of $X\otimes Y$. I am assuming $X,Y$ are nonempty. $\endgroup$
    – Tyrone
    Commented Mar 8, 2023 at 10:54
  • $\begingroup$ Then doesn't this tell us $\theta$ is a bijection? Also, when you say "the above observation", it's not clear which sentence you mean. It could be anything prior in the post, because it's not signposted $\endgroup$
    – David Roberts ♦
    Commented Mar 8, 2023 at 12:56
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    $\begingroup$ I think $\mathcal{H}$ has only finite coproducts. I agree that the statement you are referring to in the note is poorly worded. I didn't mean to mislead. However, I'm not sure I follow your comment about the inclusion of $KHaus$. If $Y$ is compact and $X$ is not, then there is no proper map $X\rightarrow Y$. $\endgroup$
    – Tyrone
    Commented Mar 8, 2023 at 15:06
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    $\begingroup$ But these definitions do not yield an adjunction because if $K$ is compact and $X$ is noncompact, then $\mathcal{H}(X,K)$ is empty while $\mathcal{K}(X_\infty,K)$ is not. $\endgroup$
    – Tyrone
    Commented Mar 8, 2023 at 18:16
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    $\begingroup$ What you can do is this: let $\mathcal{K}^*$ be the category with objects pointed compact $T_2$ spaces. A morphism in $\mathcal{K}^*$ is a map $f:K\rightarrow L$ such that $f^{-1}(\ast_L)=\{\ast_K\}$. One-point compactification determines an adjunction $\mathcal{H}\dashv\mathcal{K}^*$. The right adjoint deletes the basepoint. Both categories have finite coproducts: in $\mathcal{H}$ these are disjoint unions and in $\mathcal{K}^*$ they are wedge sums. Both categories have monoidal products (Tychonoff vs. smash). Both finite coproducts and monoidal products are preserved by the adjunction. $\endgroup$
    – Tyrone
    Commented Mar 8, 2023 at 18:18

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