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Let $X$ be a topological space. The endofunctor $\_\times X$ of the category of all topological spaces does in general not possess a right adjoint, since the category is not cartesian closed.

  1. Is it nevertheless true in general, that $colim(A_i)\times X\cong colim(A_i\times X)$?
  2. If 1) is false: Is it at least true in full generality for colimits indexed over the natural numbers $...\rightarrow X_i\rightarrow X_{i+1}\rightarrow ...$?
  3. If 2) is false: What are topological conditions on $X_i$ and $A$, such that 2) gets right?
Newuser
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  • Welcome to Math.SE! Please include your own ideas about solving this question, so it is easier for other people to help you. – Hrodelbert Apr 28 '15 at 10:00
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    On (3): Well, it suffices for $(-) \times X$ to have a right adjoint, so you can assume $X$ is locally compact Hausdorff. – Zhen Lin Apr 28 '15 at 10:41
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    Locally compactness is sufficient, too. One does not need to impose the Hausdorff condition. – Newuser Apr 28 '15 at 12:33

1 Answers1

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Here is a counterexample for the directed limit:
Let $X_n$ be a wedge of $n$ circles $C_1,\dots,C_n$. We have a sequence of closed inclusions $X_1\hookrightarrow X_2\hookrightarrow\dots$ whose colimit is $X$, a wedge of countably many circles. There is a continuous bijection $j:\text{colim}(X_n\times\Bbb Q)\to X\times\Bbb Q$ which I claim is not a homeomorphism. Namely, if $A$ is the subset of $\text{colim}(X_n\times\Bbb Q)$ whose intersection with the cylinder $C_n\times\Bbb Q$ is $$ \left\{ (e^{2\pi i x},y)\ \middle| \ \frac\pi n \le y \le \frac\pi n +\min(x,1-x)\right\} $$ then $A$ is closed, but $j(A)$ is not closed in $X\times\Bbb Q$ as $(0,0)$ is a limit point of $j(A)$.

We would have a homeomorphism, though, if $Y$ (which in the example was $\Bbb Q$) were locally compact, as that would make the functor $(-)\times Y$ a left adjoint to the functor $(-)^Y$, and left adjoints preserve all colimits.

It also works, for any $Y$, if the diagram is such that we can choose a subset $\cal S$ of the spaces in the diagram $X:\mathscr J\to\mathbf{Top}$ such that

  • every space in the diagram is either in $\cal S$ or has a map to some space in $\cal S$
  • the quotient map $\coprod_{i\in\mathscr J} X_i\to\text{colim}(X_i)$ restricts to a perfect map $\coprod_{s\in\mathcal S}X_s\to\text{colim}(X_i)$ (Note that the previous point implies that this restriction is a quotient map).

This is because for every perfect map $p:X\to Y$, the product $p\times\mathbf 1_Z:X\times Z\to Y\times Z$ is closed, thus a quotient map.

Stefan Hamcke
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  • Does the situation get better if one restricts his self to spaces in a convenient category like CG, CGWH or CGH? – Tom Apr 29 '15 at 08:31
  • @Tom: Yes, that's actually the motivation for a convenient category of spaces. One wants it to be Cartesian closed. In such a category, the functor $(-)×Y$ is left adjoint to the exponential $(-)^Y$. another consequence is that products of quotient maps are quotient maps. This fails in $\mathbf{Top}$ if $Y$ is not locally compact. – Stefan Hamcke Apr 29 '15 at 15:00
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    @Tom: Of course this means that the product is not longer the usual one. You have to change the product topology in order to make $X\times Y$ another object in your convenient category. This is not as bad as it sounds. For example, the product of CW complexes in CGWH is again a CW complex. – Stefan Hamcke Apr 29 '15 at 15:04