Today I Learned

March 8th

Today I learned some more solid motivation for the compact-open topology, again from the nlab page . In particular, this time compactness actually matters. The following is our claim.

What's remarkable here is that $X,Y,Z$ have no hypotheses other than being topological spaces. In particular, we don't need locally compact or even something like Hausdorff.

We begin by showing that $\op{curry}$ is actually well-defined.

Fix any $U\subseteq Z$ open so that we want to show its preimage\[\op{curry}(x)^{-1}(U)=\{y\in Y:f((x,y))\in U\}\subseteq U\]is also open. However, we know that $f^{-1}(U)\subseteq X\times Y$ is open because $f$ is continuous, and the idea is to attempt to project $f^{-1}(U)$ onto $\{x\}\subseteq X.$

Because $X\times Y$ has basis pairs of open sets, we know that any $y\in Y$ with $(x,y)\in f^{-1}(U)$ has some $U_\bullet$ and $U_y$ with $(x,y)\in U_\bullet\times U_y\subseteq f^{-1}(U).$ Further, we note that each $U_y$ has $U_y\subseteq\op{curry}(x)^{-1}(U)$ because\[\op{curry}(x)(U_y)=f(\{x\}\times U_y)\subseteq U.\]It follows that we can write\[\op{curry}(x)^{-1}(U)\subseteq\bigcup_{(x,y)\in f^{-1}(U)}U_y\subseteq\op{curry}(x)^{-1}(U).\]We see that equalities must hold, and the middle arbitrary union is an open set, so $\op{curry}(x)^{-1}(U)$ is therefore open. Thus, $\op{curry}(x)$ is continuous, which is what we wanted. $\blacksquare$

Anyways, we now go after the proposition directly. We need to show that each open set $U$ in $\op{Hom}(Y,Z)$ has $\op{curry}^{-1}(U)$ open in $X.$ We'll have to use the compact-open topology sometime, so we begin by reducing to subbasis elements. Letting $V(K,U):=\{f:f(K)\subseteq U\}$ being our subbasis elements of $\op{Hom}(Y,Z),$ we note that we can write\[U=\bigcup_{\alpha\in\lambda}\left(\bigcap_{k=1}^{n_\alpha}V(K_{\alpha,k},U_{\alpha,k})\right),\]for $V_{\bullet,\bullet}\subseteq Y$ compact and $U_{\bullet,\bullet}\subseteq Z$ open. Now, we note that\[\op{curry}^{-1}(U)=\bigcup_{\alpha\in\lambda}\left(\bigcap_{k=1}^{n_\alpha}\op{curry}^{-1}\big(V(K_{\alpha,k},U_{\alpha,k})\big)\right),\]so it suffices to show that $\op{curry}^{-1}(V(K,U))$ is open for any $K$ and $U.$

As an aside, we note briefly that it is in fact true that, for any function $\varphi:S\to T$ and subsets $\{T_\alpha\}_{\alpha\in\lambda}$ of $T,$\[\varphi^{-1}\left(\bigcup_{\alpha\in\lambda}T_\alpha\right)=\bigcup_{\alpha\in\lambda}\varphi^{-1}(T_\alpha),\]for $s$ is in either set if and only there exists an $\alpha\in\lambda$ for which $\varphi(s)\in T_\alpha.$ Similarly,\[\varphi^{-1}\left(\bigcap_{\alpha\in\lambda}T_\alpha\right)=\bigcap_{\alpha\in\lambda}\varphi^{-1}(T_\alpha),\]for $s$ is in either set if and only if $\varphi(s)\in T_\alpha$ for each $\alpha\in\lambda.$ We used these facts implicitly to reduce to subbasis; I only mention this because I thought the $\bigcap$ equation above was actually false because $\varphi(S_1\cap S_2)$ is not necessarily equal to $\varphi(S_1)\cap g(S_2)$ for subsets $S_1,S_2\subseteq S.$

Continuing, we fix $K_Y\subseteq Y$ compact and $U_Z\subseteq Z$ open, and we want to show $\op{curry}^{-1}(V(K_Y,U_Z))=:S$ is open. Expanding out the definitions, we want to show that\[S=\{x\in X:[y\mapsto f((x,y))]\in V(K_Y,U_Z)\}=\{x\in X:f(\{x\}\times K_Y)\subseteq U_Z\}\]is open. Here we see opportunity to use the continuity of $f,$ so we take it: it suffices to show that\[S=\left\{x\in X:\{x\}\times K_Y\subseteq f^{-1}(U_Z)\right\}\]is open. The set $f^{-1}(U_Z)$ is pretty much just an arbitrary open set by continuity, so we let it be $U_{XY}\subseteq X\times Y.$

To continue the argument, we look locally. It suffices to show that, for each $x\in S,$ there is an open neighborhood $U_x\subseteq X$ around $x$ for which $U_x\subseteq S.$ Indeed, then we could write\[S\subseteq\bigcup_{x\in S}U_x\subseteq S,\]from which $S$ being open would follow. So we have to show that $\{x\}\times K_Y\subseteq U_{XY}$ implies there exists an open set $U_x\subseteq X$ containing $x$ for which $U_x\times K_Y\subseteq U_{XY}.$ Visually, we have to "expand'' $\{x\}$ into a slightly larger box inside $U_{XY}.$

We should have to use the compactness of $K_Y$ eventually because I promised it was important, so we attempt to build an open cover of $K_Y.$ For each $k\in K_Y,$ we note that $(x,k)\in U_{XY}.$ Using the basis of $X\times Y,$ this gives us open sets $U_{(x,k),x}$ and $U_{(x,k),k}$ for which\[(x,k)\in U_{(x,k),x}\times U_{(x,k),k}\subseteq U_{XY}.\]We would like to just intersect all of the $U_{(x,k),x}$ to generate a set around $x$ whose product with $K_Y$ is inside of $U_{XY}.$ However, we can't take arbitrary intersections, so we use the compactness of $K_Y$ to make it a finite intersection. Because the $U_{(x,k),k}$ are neighborhoods around each $k\in K_Y,$ we see\[K_Y\subseteq\bigcup_{k\in K_Y}U_{(x,k),k},\]so we have an open cover. So we extract $k_1,\ldots,k_n\in K_Y$ for which $U_{(x,k_\bullet),k_\bullet}$ completely coves $K_Y$ by compactness, and we claim that\[U_x:=\bigcap_{\ell=1}^nU_{(x,k_\ell),x}\]works. Indeed, $U_x$ is the finite intersection of open sets all containing $x,$ so $U_x$ is also an open set containing $x.$ And for any $x'\in U_x,$ we have that\[\{x'\}\times U_{(x,k_\bullet),k_\bullet}\subseteq U_{(x,k_\bullet),x}\times U_{(x,k_\bullet),k_\bullet}\subseteq U\]for each $k_\bullet,$ so $\{x'\}\times K_Y\subseteq U$ follows. Thus, we do have $U_x\subseteq U_{XY},$ and we are done here. $\blacksquare$

As a final remark, we note that both the "open'' and the "compact'' of the compact-open topology played subtle but crucial roles. The "open'' part was used in the continuity of $f$ to get $U_{XY}=f^{-1}(U_Z)$ open. This feels quite natural to me, for it's not clear how we would use the continuity of $f$ except for requiring the open set to be open.

The "compact'' part was used at the very end to expand out $\{x\}\times K_Y$ to an open neighborhood around $\{x\}.$ I can almost believe that compactness is the "correct'' hypothesis to make currying work because otherwise it's not clear if we're always able to expand out $\{x\}\times K_Y.$ Namely, some kind of smallness condition is necessary on $K_Y$ to be able to expand out a box like this.