4
$\begingroup$

Consider the following two exercises in May's A Concise Course in Algebraic Topology.

  1. (a) Any subspace of a weak Hausdorff space is weak Hausdorff. (b) Any closed subspace of a $k$-space is a $k$-space. (c) An open subset $U$ of a compactly generated space $X$ is compactly generated if each point has an open neighborhood in $X$ with closure contained in $U$.

  2. A Tychonoff (or completely regular) space $X$ is a $T_1$-space (points are closed) such that for each point $x \in X$ and each closed subset $A$ such that $x \notin A$, there is a function $f: X \to I$ such that $f(x) = 0$ and $f(a) = 1$ if $a \in A$. Prove the following. (a) A space is Tychonoff if and only if it can be embedded in a cube (a product of copies of $I$). (b) There are Tychonoff spaces that are not $k$-spaces, but every cube is a compact Hausdorff space.

In view of these two problems, what should one mean by a "subspace" of a compactly generated space? We do not want to restrict the allowable subsets.

$\endgroup$
1
$\begingroup$

The most naïve idea is to define the subspace topology to be the k-ified relative topology, so it is no longer necessarily induced from the ambient space simply from intersections. But this doesn't really invoke the previous problems.

edit I think this is what May has in mind, since the k-ified subspace topology satisfies the relevant universal property; see the bottom of p.155 of this appendix.

$\endgroup$
1
$\begingroup$

To elaborate a bit on Owen's answer, the reason the subspace topology on a subset $A\subset X$ is normally defined the way it is is such that the following property holds:

For any space $Y$, a function $f:Y\to A$ is continuous iff the composition $if:Y\to X$ is continuous, where $i:A\to X$ is the inclusion. That is, continuous maps to $A$ are exactly continuous maps to $X$ whose images are contained in $A$.

In the category of compactly generated (weak Hausdorff) spaces, this property holds if you equip $A$ with the $k$-ification of the usual subspace topology. Indeed, this is immediate from the universal property of $k$-ification: if $A$ denotes $A$ with the usual subspace topology and $kA$ denotes its $k$-ification, then for any compactly generated space $Y$, a map $Y\to A$ is continuous iff it is continuous as a map to $kA$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy