2
$\begingroup$

We start with some preliminary definitions (necessary because there is not much literature on this):

  • a test map is a continuous function $\varphi:V\rightarrow X$ where $V$ is an open subspace of $\mathbb{R}^n$ for some $n$.
  • Given a topological space $X$, we say that a subset $U$ of $X$ is numerically open if $\varphi^{-1}(U)$ is open for every test map $\varphi$.
  • A topological space $X$ is said to be numerically generated if a subset $U$ of $X$ is open if and only if it is numerically open.
  • Given any topological space $X$, we define the numericalization of $X$, denoted $X^\#$, to be the set underlying $X$ equipped with the final topology induced by the family of test maps into $X$.
  • The numerically generated product of two numerically generated spaces $X$ and $Y$ is the numericalization of the product space $X\times Y$ (henceforth simply denoted $X\times Y$).
  • The numerically generated mapping space of the numerically generated spaces $X$ and $Y$, denoted $Map(X,Y)$, is the collection of all continuous maps from $X$ to $Y$ equipped with the numericalization of the compact-open topology.
  • The category $\mathbf{Num}$ is the full subcategory of $\mathbf{Top}$ with objects as numerically generated spaces; this has as coproduct the disjoint union with the disjoint union topology (it is not difficult to show that the disjoint union of numerically generated spaces is numerically generated) and product as the numerically generated product.

I wish to show that $\mathbf{Num}$ is a cartesian-closed category. Ultimately, this simply boils down to showing that the numerically generated mapping space is an exponential object in $\mathbf{Num}$.

Naturally, I would want to use the typical bijection $h: Hom(X\times Y,Z)\rightarrow Hom(X,Z^Y)$ given by currying, but it remains to show that given a continuous map $f:X\times Y\rightarrow Z$, that the map $g:X\rightarrow Map(Y,Z)$ defined by $g(x)=f(x,-)$ is continuous (as well as that $f(x,-)$ is continuous too). However, I'm tripping up on showing this.

$\endgroup$
  • $\begingroup$ Have you tried the emulate the usual arguments that go into showing that this is true for compactly generated spaces? $\endgroup$ – Piotr Pstrągowski Mar 14 '14 at 18:11
  • $\begingroup$ @PiotrPstragowski I have not, since I don't know the usual arguments; this is my first introduction to algebraic topology and we're not using compactly generated spaces, but I'll try to take a look at the proof of that subcategory. $\endgroup$ – Hayden Mar 14 '14 at 18:17
  • $\begingroup$ I've never seen this definition before! Where did you encounter it? $\endgroup$ – Zhen Lin Mar 14 '14 at 19:57
  • $\begingroup$ Professor Barwick at MIT; we're studying the Fundamental Groupoid & the Postnikov Tower in our Topology Seminar. Having done some searching before, slight mention of the topic is seen in some higher level algebraic topology papers, but nothing more than that. $\endgroup$ – Hayden Mar 14 '14 at 20:00
  • $\begingroup$ There is quite a lot of literature in this area. See for example Booth, P.I., Tillotson, J., "Monoidal closed categories and convenient categories of topological spaces". Pacific J. Math. 88 (1980) 33-53. See also Section 5.9 of my book "Topology and Groupoids". It is more usual to take the test maps from some compact Hausdorff spaces, so it is interesting to see this definition. $\endgroup$ – Ronnie Brown Mar 15 '14 at 12:47

Your Answer

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

Browse other questions tagged or ask your own question.