2

There are various incompatible definitions of $k$-space/compactly generated space in the literature, as explained in wikipedia and many posts on this site. But let's focus on Definition 2 from the wikipedia article, which is the one commonly used in algebraic topology.

A space $X$ is called compactly generated (CG2) if its topology is determined by continuous maps from arbitrary compact Hausdorff spaces. In other words, it is a space such that a subset $A\subseteq X$ is closed if it is $k$-closed in the sense that $u^{-1}[A]$ is closed in $K$ for every compact Hausdorff space $K$ and every continuous map $u:K\to X$.

Is that the same as requiring that the topology on $X$ be coherent with the family of images of compact Hausdorff spaces under continuous maps?

And at the level of sets, given a topological space $X$, are these two conditions on a subset $A\subseteq X$ equivalent?

  • (1) $u^{-1}[A]$ is closed in $K$ for every compact Hausdorff space $K$ and every continuous map $u:K\to X$.
  • (2) $A\cap u[K]$ is closed in $u[K]$ for every compact Hausdorff space $K$ and every continuous map $u:K\to X$.

Clearly (2) implies (1). Can you have (1) without (2)?

PatrickR
  • 7,165
  • Might be good to clarify that "At the level of sets..." etc. is a rephrasing of the same question, not a separate question. – M W Jan 18 '24 at 07:27
  • @MW Well, it's close but not an exact rephrasing. If (1) and (2) were equivalent, then the two notions of $k$-closed would be equivalent for sets, and the corresponding notions of $k$-space would also be equivalent. But one could imagine (1) and (2) are not equivalent for a specific set, however if we quantify over all subsets of $X$ it could still be the case that the two notions of $k$-space are equivalent. As it turns out from your answer, the two notions are not equivalent. – PatrickR Jan 18 '24 at 19:31
  • 1
    Let $\tau_{\mathcal I}$ be the final topology for the family $\mathcal I$ of all inclusions $i\colon u[K]\to X$, where $K$ is compact Hausdorff and $u\colon K\to X$ is continuous. Crucially (exercise:) ), $v\colon K\to X$ is continuous if and only if it is $\tau_{\mathcal I}$ continuous. Thus once there is a single set $A$ where we have (1) and not (2), then $\tau_{\mathcal I}$ is coherent with images of compact Hausdorff spaces, but not final with respect to maps from compact Hausdorff spaces, so $(X,\tau_{\mathcal I})$ is the counterexample for the notions of $k$-space being equivalent. – M W Jan 18 '24 at 20:41
  • To elaborate a bit more, if $S\subset X$ is not $\tau_I$ closed, then $S\cap u[K]$ is not closed in $u[K]$ for some $u[K]$. Thus $(S\cap u[K])\cap u[K]$ is not closed in $u[K]$, so $S\cap u[K]$ is not $\tau_I$-closed. Therefore, since such $u$ is also $\tau_I$-continuous as well (exercise, or maybe further comment), $\tau_I$ is always coherent with the continuous images of compact Hausdorffs. But it will not be final with respect to these maps, since the set $A$ that satisfies 1 and not 2 will be closed in the final topology, but not $\tau_I$ closed. – M W Jan 18 '24 at 21:15
  • Thanks for the detailed explanation that $(X,\tau_I)$ would be an example where the two notions of $k$-space are not equivalent, given that there is a set $A$ in the original space $(X,\tau)$ satisfying (1) and not (2). – PatrickR Jan 19 '24 at 02:14

1 Answers1

2

The one point compactification of the Arens-Fort space seems to be a counter-example. It is not $CG_2$, but it is itself the image of a map from $\omega+1$, so certainly its topology is coherent with the images of maps from compact Hausdorff spaces, as these images include the space itself.

(To define such a map, simply let $x_n$ be any enumeration of the Arens-Fort space, and define $f(n)=x_n$, $f(\omega)=\infty$).

Remark.

To unwind this explicitly at the level of sets, if $Y$ is the Arens-Fort space and $X=Y^*$ its one-point compactification, we can see that the set $A=Y^*\backslash \{(0,0)\}$ is closed in the sense of (1), but not (2). To see this, note that if $u\colon K\to Y^*$ is a map from compact Hausdorff $K$, then $u^{-1}(\{(0,0)\})$ and $u^{-1}(\{\infty\})$ are disjoint closed sets, so by normality there is a neighborhood $U$ of $u^{-1}(\{(0,0)\})$ with $\overline{U}\cap u^{-1}(\{\infty\})=\emptyset$, hence $u(\overline{U})$ is a compact subset of $Y$, and therefore finite, as $Y$ is anticompact.

In particular, $u(\overline{U})$ is discrete, so $u^{-1}(\{(0,0)\})$ is open in $\overline{U}$ (either it is the pre-image of an isolated point in $u(\overline{U})$, or it is empty), hence open in $U$, hence open in $K$ (since $U$ is open), so $Y^*\backslash \{(0,0)\}$ is closed in the sense of (1). But since $Y^*$ is the image of a compact Hausdorff space, and $(0,0)$ is not isolated in $Y^*$, $Y^*\backslash \{(0,0)\}$ is not closed in the sense of (2).

(This is all really just unpacking the argument from this answer.)

M W
  • 10,329
  • 1
    Very nice. More generally, if $X$ (Arens-Fort space in this case) is any anticompact $T_1$ non-discrete space, its usual one-point compactification $Y=X\cup{\infty}$ is not CG2, but is coherent with the images of maps from compact Hausdorff space. Because the Fort topology on $Y$ (one-point compactification of a discrete space) is Hausdorff and is a finer topology. – PatrickR Jan 18 '24 at 19:24
  • Feel free to add an explicit example of set $A\subseteq X$ that is $k$-closed in the sense discussed but not in the sense of CG2. – PatrickR Jan 18 '24 at 19:34
  • For the comment above the above, I meant to say the Fort topology is compact Hausdorff ... – PatrickR Jan 18 '24 at 19:35
  • Thanks for the Remark. One could add a sentence mentioning the trivial case when $u^{-1}({(0,0)})=\emptyset$. And I guess $X$ is the Arens-Fort space and $Z$ is meant to be $K$? – PatrickR Jan 19 '24 at 05:37
  • @PatrickR Oh yeah $Z$ was indeed meant to be $K$. Also I guess using $X$ for the Fort-Space was pretty confusing for consistency with the original question's notation, so I changed it to $Y$ to hopefully clarify that $X=Y^*$, so $X$ and $A$ now align with how the question was asked. I might be being slow but I'm not quite following what needs to be said about the trivial case when $u^{-1}({0,0})=\emptyset$. Is there a reason that needs to be treated differently? – M W Jan 19 '24 at 08:23
  • I guess I see the issue, I talked about $(0,0)$ being isolated in $u(\overline{U})$, which isn’t always true. I rephrased. – M W Jan 19 '24 at 17:48