From Product Maps to Function-Space Maps
A map \(H:Z\times X\to Y\) can be described one \(z\in Z\) at a time: fix \(z\), and consider the function \(x\mapsto H(z,x)\). The exponential law makes this description precise. It says that, with the compact-open topology on the space of continuous functions from \(X\) to \(Y\), continuity of \(H\) is equivalent to continuity of the resulting map from \(Z\) into that function space, provided \(X\) is locally compact Hausdorff.
All spaces in this tutorial have specified topologies. The product \(Z\times X\) has the product topology. Write \(C(X,Y)\) for the set of continuous maps from \(X\) to \(Y\), equipped with the compact-open topology.
For any function \(H:Z\times X\to Y\), its adjoint is the function \(H^\#:Z\to C(X,Y)\) given by \(H^\#(z)(x)=H(z,x)\), when each slice \(x\mapsto H(z,x)\) is continuous. In particular, if \(H\) is continuous, its slices are continuous: the map \(x\mapsto(z,x)\) is continuous, and composing it with \(H\) gives the slice.
The Two Directions of the Law
We use the convention that locally compact Hausdorff means that for every \(x\in X\) and every open neighborhood \(V\) of \(x\), there is an open neighborhood \(W\) of \(x\) whose closure is compact and contained in \(V\). A compact neighborhood \(K\) of \(x\) is a compact set containing an open neighborhood of \(x\).
Proof. First suppose \(H:Z\times X\to Y\) is continuous. We prove that \(H^\#\) is continuous by checking the subbasis. Fix a compact \(K\subseteq X\) and an open \(U\subseteq Y\), and suppose \(z_0\in (H^\#)^{-1}([K,U])\). This means \(H(\{z_0\}\times K)\subseteq U\). For each \(x\in K\), continuity of \(H\) gives open neighborhoods \(A_x\) of \(z_0\) in \(Z\) and \(V_x\) of \(x\) in \(X\) such that \(H(A_x\times V_x)\subseteq U\). The sets \(V_x\) cover \(K\). Compactness gives finitely many \(x_1,\dots,x_n\) with \(K\subseteq V_{x_1}\cup\cdots\cup V_{x_n}\). The open neighborhood \(A_{x_1}\cap\cdots\cap A_{x_n}\) of \(z_0\) is then contained in \((H^\#)^{-1}([K,U])\). If \(K=\varnothing\), this preimage is all of \(Z\). Thus every such subbasic preimage is open, so \(H^\#\) is continuous.
For the reverse direction, define evaluation by \(\operatorname{ev}:C(X,Y)\times X\to Y\), \(\operatorname{ev}(f,x)=f(x)\). We first prove it is continuous under the stated local compactness and Hausdorff hypotheses. Let \(f(x)\in U\), where \(U\) is open in \(Y\). The set \(f^{-1}(U)\) is an open neighborhood of \(x\). By local compactness, choose an open neighborhood \(W\) of \(x\) with compact closure \(K=\overline{W}\subseteq f^{-1}(U)\). Then \(f(K)\subseteq U\), so \(f\in[K,U]\), and \(x\in W\). For every \(g\in[K,U]\) and \(y\in W\subseteq K\), we have \(g(y)\in U\). Hence \([K,U]\times W\) is a neighborhood of \((f,x)\) mapped into \(U\). This proves continuity of evaluation.
Now suppose \(H^\#:Z\to C(X,Y)\) is continuous. The map \(H^\#\times\operatorname{id}_X:Z\times X\to C(X,Y)\times X\), given by \((z,x)\mapsto(H^\#(z),x)\), is continuous by the definition of the product topology. Its composition with evaluation is continuous, and at \((z,x)\) its value is \(H^\#(z)(x)=H(z,x)\). Therefore \(H\) is continuous. Finally, the formulas \(H^\#(z)(x)=H(z,x)\) and \(H(z,x)=H^\#(z)(x)\) show that taking adjoints and uncurrying are inverse operations. This proves the claimed bijection. \(\square\)
The compactness of \(K\) is what makes the first direction work: finitely many product neighborhoods can be combined to give one neighborhood of \(z_0\). Local compactness is used in the other direction to find a compact set around each \(x\) on which the function remains inside \(U\).
Worked Applications
Worked Example: A Discrete Parameter Space
Let \(Z=\{z_0,z_1\}\) have the discrete topology, let \(X=[0,1]\) have its usual subspace topology from \(\mathbb{R}\), and let \(Y=\mathbb{R}\) have its standard topology. Define \(H:Z\times X\to Y\) by \(H(z_0,x)=x\) and \(H(z_1,x)=2x\). Each slice is continuous. The map \(H\) is continuous: for an open \(U\subseteq\mathbb{R}\), its preimage is the union of the open sets \(\{z_0\}\times f_0^{-1}(U)\) and \(\{z_1\}\times f_1^{-1}(U)\), where \(f_0(x)=x\) and \(f_1(x)=2x\).
Consider the compact set \(K=\{0,1\}\) and the open set \(U=(-0.5,1.5)\). For \(z_0\), the values on \(K\) are \(H(z_0,0)=0\) and \(H(z_0,1)=1\), both in \(U\). For \(z_1\), they are \(H(z_1,0)=0\) and \(H(z_1,1)=2\), and \(2\notin U\). Therefore \((H^\#)^{-1}([K,U])=\{z_0\}\), which is open in the discrete topology on \(Z\).
Check: the adjoint has \(H^\#(z_0)=f_0\) and \(H^\#(z_1)=f_1\); the compact-open test just computed gives exactly the singleton preimage \(\{z_0\}\).
Worked Example: Multiplication on the Unit Interval
Let \(Z=\mathbb{R}\), \(X=[0,1]\), and \(Y=\mathbb{R}\), with their usual topologies (and the subspace topology on \(X\)). Define \(H(z,x)=zx\). This is continuous as the restriction of the usual continuous multiplication map on \(\mathbb{R}^2\). Its adjoint is \(H^\#(z)(x)=zx\).
Take \(K=[0,1]\) and \(U=(-1,1)\). We compute the subbasic preimage. If \(|z|<1\), then for every \(x\in[0,1]\), \(|zx|=|z|x\leq |z|<1\), so \(zx\in U\). If \(|z|\geq1\), then \(x=1\) belongs to \(K\) and \(|z\cdot1|=|z|\geq1\), so \(z\notin(H^\#)^{-1}([K,U])\). Thus \((H^\#)^{-1}([K,U])=(-1,1)\), which is open in \(\mathbb{R}\).
Check: at \(z=0\), the image of \(K\) is \(\{0\}\subset U\); at \(z=1\), the value at \(x=1\) is \(1\notin U\). These outcomes agree with the computed interval.
Worked Example: A Continuous Family of Translates
Let \(Z=\mathbb{R}\), \(X=[0,1]\), and \(Y=\mathbb{R}\) with the topologies just specified. Define \(F:Z\to C(X,Y)\) by \(F(z)(x)=z+x\). For \(K=[0,1]\) and \(U=(0,3)\), the condition \(F(z)\in[K,U]\) says that every value \(z+x\), for \(0\leq x\leq1\), lies strictly between \(0\) and \(3\). The smallest value is \(z\), attained at \(x=0\), so we need \(z>0\). The largest is \(z+1\), attained at \(x=1\), so we need \(z+1<3\), or \(z<2\). Consequently \(F^{-1}([K,U])=(0,2)\), an open set.
More generally, \(F\) is continuous by the theorem: its uncurried map is \(H(z,x)=z+x\), which is continuous on \(\mathbb{R}\times[0,1]\). Check: at \(z=1\), the image of \(K\) is \([1,2]\subset(0,3)\); at \(z=0\), the value at \(x=0\) is \(0\notin(0,3)\); at \(z=2\), the value at \(x=1\) is \(3\notin(0,3)\). Both endpoint exclusions match the strict inequalities.
Why the Domain Hypothesis Matters
The forward implication does not need local compactness: compactness of the sets \(K\) in the compact-open subbasis is enough for the finite-cover argument. Local compactness enters when continuity of a function-space map must be converted back into continuity on the product. Here is a concrete failure of that conversion when the domain is not locally compact.
Let \(X=\mathbb{Q}\) with the subspace topology inherited from \(\mathbb{R}\). This space is not locally compact. Indeed, if a compact \(K\subseteq\mathbb{Q}\) contained an open neighborhood of some rational \(x\), then \(K\), being compact in the Hausdorff space \(\mathbb{R}\), would be closed in \(\mathbb{R}\). It would contain a real interval: it contains all rationals in some interval around \(x\), and those rationals are dense there. But a closed subset of \(\mathbb{R}\) containing that interval contains its irrational points too, contradicting \(K\subseteq\mathbb{Q}\).
Give \(Y=\{0,1\}\) the discrete topology, and give \(C(X,Y)\) the compact-open topology. Fix \(x\in\mathbb{Q}\), and let \(f\) be the constant-zero function. We claim evaluation is not continuous at \((f,x)\). A basic neighborhood of \(f\) is a finite intersection of subbasic sets \([K_j,U_j]\) containing \(f\). Each \(U_j\) must contain \(0\), so it is either \(\{0\}\) or \(Y\). Let \(K\) be the union of those \(K_j\) for which \(U_j=\{0\}\). This is compact, and any function vanishing on \(K\) belongs to the chosen finite intersection.
Given any open neighborhood \(O\) of \(x\) in \(\mathbb{Q}\), we cannot have \(O\subseteq K\), since then \(K\) would be a compact neighborhood of \(x\). Choose \(y\in O\setminus K\). Compact subsets of the Hausdorff space \(\mathbb{Q}\) are closed, so \(O\setminus K\) is open in \(\mathbb{Q}\). It contains a nonempty set \(W=(a,b)\cap\mathbb{Q}\) around \(y\), where \(a,b\) are irrational and the interval is chosen small enough that \(W\subseteq O\setminus K\). This \(W\) is also closed in \(\mathbb{Q}\), because its boundary points \(a,b\) are not in \(\mathbb{Q}\). The function \(g:\mathbb{Q}\to Y\) that equals \(1\) on \(W\) and \(0\) elsewhere is continuous: the preimages of both points of the discrete space \(Y\) are clopen. It vanishes on \(K\), so it lies in the chosen neighborhood of \(f\), but \(g(y)=1\).
Thus every product neighborhood of \((f,x)\) contains some \((g,y)\) with evaluation equal to \(1\), even though \(f(x)=0\). The preimage of the open set \(\{0\}\) under evaluation is not a neighborhood of \((f,x)\), so evaluation is not continuous there. The exponential law therefore cannot be asserted for arbitrary \(X\).
Common pitfall. The compact-open topology is designed to control a function on compact subsets, but a compact set need not contain any neighborhood of a given point. In the proof of continuity of evaluation, it is not enough to choose an arbitrary compact \(K\) containing \(x\); one needs a compact neighborhood whose interior contains \(x\) and on which the function stays in the required open set. Local compactness supplies precisely this.
Check Your Understanding
Use the subbasis definition and keep track of which direction of the exponential law is being applied.
- For a continuous \(H:Z\times X\to Y\), explain where compactness of \(K\) is used to prove that \((H^\#)^{-1}([K,U])\) is open.
- In the multiplication example, compute \((H^\#)^{-1}([0,1],(-2,2))\).
- For \(F(z)(x)=z+x\), compute \(F^{-1}([[0,1],(1,4)])\), taking account of both endpoints of \([0,1]\).
- Which direction of the exponential law uses local compactness, and what role does the compact neighborhood play?
- Why does the proof that evaluation is continuous require a neighborhood of \(x\) inside the compact set \(K\), rather than only \(x\in K\)?