Today I Learned

March 6th

Today I learned a little motivation for the compact-open topology: the evaluation map is continuous, from the nlab page . That is, we have the following proposition.

Here, $\op{Hom}(X,Y)$ means the set of continuous functions $X\to Y,$ equipped with the compact-open topology. Also, we are ordering the arguments like this to have a future with currying. However, I did not do currying today.

Fix $U_Y\subseteq Y$ any open set. We need to show that its preimage under $\op{ev}$ is open. We note that it suffices to show, for each $(x,f)\in\op{ev}^{-1}(U_Y),$ that there is an open neighborhood $U_{(x,f)}$ around $(x,f)$ such that $U_{(x,f)}\subseteq\op{ev}^{-1}(U_Y).$ Indeed, we could then write\[\op{ev}^{-1}(U_Y)=\bigcup_{(x,f)\in\op{ev}^{-1}(U_Y)}U_{(x,f)},\]which is manifestly open. In fact, this condition is equivalent to $\op{ev}^{-1}(U_Y)$ being open, for it if were open, then we could just set $U_{(x,f)}=\op{ev}^{-1}(U_Y).$ Anyways, the point is that we are able to look locally at points $(x,f)$ of $X\times\op{Hom}(X,Y).$

It remains to exhibit $U_{(x,f)}.$ This comes down to the interaction of $X$ being locally compact with the compact-open topology. For any function $f,$ we can use the continuity of $f$ to yield an open set $f^{-1}(U_Y),$ which in fact contains $x.$ The reason we do this is to use local compactness: now we get an open set $U_x$ and compact $V$ such that\[x\in U_x\subseteq V\subseteq f^{-1}(U_Y).\]Read this definition of local compactness as essentially meaning that we have a basis of compact sets around $x.$

We still have to use the compact-open topology. Well, we have an instantiated a compact $V\subseteq X$ in our input space, and we have an open set $U_Y\subseteq Y$ of the output space, so we note that\[\{g\in\op{Hom}(X,Y):g(V)\subseteq U_Y\}\]is open in $\op{Hom}(X,Y).$ So we claim that\[U_{(x,f)}:=U_x\times\{g\in\op{Hom}(X,Y):g(V)\subseteq U_Y\}\]suffices. Certainly this open in the product topology. Further, $x\in U_x$ and $f(V)\subseteq f(f^{-1}(U_Y))\subseteq U_Y,$ so $U_{(x,f)}$ is indeed a neighborhood of $(x,f).$ And finally,\[\op{ev}(U_{(x,f)})=\bigcup_{g(V)\subseteq U_Y}g(U_x)\subseteq\bigcup_{g(V)\subseteq U_Y}g(V)\subseteq U_Y,\]which is what we wanted. So we have exhibited our local neighborhoods, and we are done. $\blacksquare$

Of course, this is not really motivation for the compact-open topology. We could replace "locally compact'' with something like "locally open'' and the topology on $\op{Hom}(X,Y)$ with an "open-open'' topology, and the proof would still work fine.

However, the fact that w are focusing on mostly locally compact topologies in what I'm reading (e.g., for Pontryagin duality) motivates the "locally compact'' condition for me, which makes the compact-open topology feel more natural given the above argument.