13

It has been a while since I studied algebraic topology, and I wanted to revisit homotopy theory. Determined to take a more sustainable approach, I started by questioning and verifying every result in one of my books, Switzer's Algebraic Topology - Homology and Homotopy.

So in the preliminaries, the compact-open topology on the set $Y^X$ of functions $f: X \to Y$ is defined to be generated by:

$$N_{U,K} = \{f: X \to Y \mid f(K) \subseteq U\},\quad U \subseteq Y \text{ open}, K \subseteq X \text{ compact}$$

One of the principal properties of a topology on $Y^X$ would be that it makes the evaluation mapping $e: Y^X \times X \to Y, e(f,x) = f(x)$ continuous (I know that this applies to the COT only under extra conditions, notably if $X$ is locally compact).

So, for $U \subseteq Y$ open we expect $e^{-1}(U)$ to be open in the product topology. This amounts to, for every $x$ with $f(x) \in U$ for some $f$, the existence of a neighborhood $V_x$ such that there is a neighborhood $T$ in $Y^X$ with $f(V_x) \subseteq U$ for each $f \in T$.

However, taking such $T$s as generators -- explicitly:

$$T_{U,V} = \{f \mid f(V) \subseteq U\} \quad V \subseteq X, U \subseteq Y \text{ open}$$

gives rise to a different topology than the compact-open topology. So what compelling reasons are there to consider the compact-open topology rather than the one I just described? If applicable, historical references are also appreciated.

In particular, I'm interested in results where we can see that the properties of the COT are really "needed" for the proof to follow through.

Lord_Farin
  • 17,924
  • 9
  • 52
  • 132
  • See the "properties" section of the wikipedia page: https://en.wikipedia.org/wiki/Compact-open_topology#Properties – Thomas Andrews Jan 04 '15 at 21:58
  • 1
    In particular, in the compact-open topology, evaluation is not necessarily continuous, unless $X$ is "nice" - say, locally compact and Hausdorff. – Thomas Andrews Jan 04 '15 at 22:00
  • @ThomasAndrews Well, thanks, I already read that. But that page doesn't describe why one would not choose the topology I mentioned. – Lord_Farin Jan 04 '15 at 22:01
  • I took "one of the principal properties" to mean you thought the compact-open topology had this property, which you would know it did not in most cases. Separately, it's entirely unclear in your "definition" what your subbase for this new topology is. What $T$ exactly are you allowed to choose? – Thomas Andrews Jan 04 '15 at 22:20
  • @Thomas You're right to point out the flaws in my question; I've edited to try and mitigate them. – Lord_Farin Jan 04 '15 at 22:27
  • I think this definition guarantees continuity of $e$ but it's too fine to give a natural topology. Even some sequences of functions that converge uniformly won't converge under this topology. – subrosar Jan 08 '21 at 23:27
  • There is a related thread in MO https://mathoverflow.net/questions/44358/compact-open-topology?rq=1 – Ken Apr 01 '21 at 05:14
  • Also, the topology OP described has the disadvantage that a continuous map $\varphi:X\to Y$ does not necessarily induce a continuous map $\varphi^\ast :Z^Y\to Z^X$, where $Z$ is another space. – Ken Apr 01 '21 at 05:16

2 Answers2

4

For starters, the compact-open topology on $\mathscr C(X,Y)$ gives you uniform convergence on compact subsets when $Y$ is metric. The product topology on $Y^X$ gives you pointwise convergence. :)

Ted Shifrin
  • 125,228
2

In Theorem 46.11 of Munkres' book on Topology the following is shown. Suppose we take the compact-open topology on $\mathcal C(X,Y)$. If $Z$ is any space and if $f:X\times Z\to Y$ is continuous, then the map $F:Z\to \mathcal C(X,Y)$ is continuous where $F(z)$ is the map given by $x\mapsto f(x,z)$. We say that the map $F$ is induced by $f$.

Let $\mathcal F$ denote the set $\mathcal C(X,Y)$ with some topology $\tau$ which has the property that the evaluation map $e:X\times \mathcal F\to Y$ is continuous where $e(x,f)=f(x)$. By the above theorem we have that the map $E$ which is induced by $e$ is continuous. As $E:\mathcal F\to \mathcal C(X,Y)$ is given by $E(f)(x)=e(x,f)=f(x)$, it follows that $E(f)=f$ and so $E$ is the identity map. Thus, we see that $\tau$ contains the compact-open topology.

We have shown that if a topology $\tau$ on $\mathcal C(X,Y)$ makes the evaluation map continuous, then it must contain the compact-open topology. In Theorem 46.10 of Munkres' book, is shown that if $X$ is locally compact, the compact-open topology on $\mathcal C(X,Y)$ makes the evaluation map continuous. Thus, it follows that when $X$ is locally compact, the compact-open topology is the smallest topology on $\mathcal C(X,Y)$ which makes the evaluation map continuous.

So if $X$ is not locally compact, it is possible that there is a strictly finer/larger topology on $\mathcal C(X,Y)$ which makes the evaluation map continuous.

Shameek Paul
  • 51
  • I want to ask why if $E$ is identity map, then it implies that for $\tau'$ being compact-open topology, $\tau' \subseteq \tau$ ($\tau$ contains compact-open topology). IMO, since $E$ is identity map, $\therefore \mathcal{F} = E(\mathcal{F}) \subseteq Map(X, Y)$, and $\therefore \tau \subseteq \tau'$ – Kelvin Chan Oct 12 '24 at 02:15
  • @Kelvin Chan As the identity map $E:(\mathcal F,\tau)\to (\mathcal F,\tau')$ is continuous, for every open set $U\in \tau'$, we have that $U=E^{-1}(U)\in \tau$ and so it follows that $\tau'\subseteq \tau$. – Shameek Paul Oct 13 '24 at 04:42