Timeline for Does the category of locally compact Hausdorff spaces with proper maps have products?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2023 at 18:18 | comment | added | Tyrone | 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. | |
| Mar 8, 2023 at 18:16 | comment | added | Tyrone | 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. | |
| Mar 8, 2023 at 17:52 | comment | added | Oddly Asymmetric | In fact I'm fairly sure that's a proof that $\mathcal{H}$ won't have finite coproducts: if $X$ or $Y$ are both not compact, then $X \oplus Y$ cannot be compact, and so must be $X^{*} \sqcup Y^{*}$ minus a point, leaving wlog $X^{*}$ intact. Since the coprojection $\iota_{X}$ is proper, and the inclusion into the disjoint sum is an extension of it $X$ must have been compact after all. | |
| Mar 8, 2023 at 17:18 | comment | added | Oddly Asymmetric | My point is any map $X \to Y$ with $Y$ compact must have come from a compact space, so if you define $*:LKHaus \to KHaus$ by one-point compactification on non-compact objects, identity on compact objects, and by the obvious definition on morphisms (the only non-obvious case is when $X$ is compact, $Y$ is not, but then you can just post compose with the inclusion $Y \to Y^{*}$, and can check this is functorial). Hence you have $*$ is left adjoint to the inclusion, so will need to preserve coproducts. You will therefore need $(X \oplus Y)^{*}=X^{*} \sqcup Y^{*}$ but it's not clear to me how. | |
| Mar 8, 2023 at 15:06 | comment | added | Tyrone | 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 $X\rightarrow Y$. | |
| Mar 8, 2023 at 14:46 | comment | added | Oddly Asymmetric | @Tyrone, thanks for this, I was thinking a proof might go something like this. In the note Emily mentioned, you say that it does have coproducts, yet it won't be the same coproduct as in Top, because then one-point compactification wouldn't preserve it, and one-point compactification is left adjoint to the inclusion of KHaus. Could you expand on that point? | |
| Mar 8, 2023 at 14:44 | vote | accept | Oddly Asymmetric | ||
| Mar 8, 2023 at 14:30 | vote | accept | Oddly Asymmetric | ||
| Mar 8, 2023 at 14:31 | |||||
| Mar 8, 2023 at 12:56 | comment | added | David Roberts♦ | 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 | |
| Mar 8, 2023 at 11:41 | history | edited | Tyrone | CC BY-SA 4.0 |
added 59 characters in body
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| Mar 8, 2023 at 10:54 | comment | added | Tyrone | 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. | |
| Mar 8, 2023 at 9:31 | history | answered | Tyrone | CC BY-SA 4.0 |