The map $(x,z)\mapsto x\cdot z$ on $\mathbb{R}\times \mathbb{R}$ and its transpose $z\mapsto (x\mapsto x\cdot z)$ traced through the exponential law

Take $X:=\mathbb{R}$, $Z:=\mathbb{R}$ and $Y:=\mathbb{R}$, all with the usual metric (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded), and let

$f:\mathbb{R}\times \mathbb{R}\to \mathbb{R},f(x,z):=x\cdot z\mathrm{.}$

Write $F:=\mathrm{\Phi }(f)$ for its transpose, so $F(z)(x)=x\cdot z$: each $F(z)$ is the multiplication-by-$z$ map of $\mathbb{R}$. This example checks every clause of the exponential law (The exponential law: for a locally compact metric $X$ and any spaces $Z$ and $Y$, transposition is a bijection between $C(X\times Z,Y)$ and $C(Z,C(X,Y))$ with the compact-open topology) by hand on this pair:

1. $f$ is continuous on $\mathbb{R}\times \mathbb{R}$ with the product topology (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space);
2. each $F(z)$ is continuous, being Lipschitz with constant $\mathrm{\mid }z\mathrm{\mid }$, so $F(z)\in C(\mathbb{R},\mathbb{R})$;
3. $F:\mathbb{R}\to C(\mathbb{R},\mathbb{R})$ is continuous for the compact-open topology, directly: if $\mathrm{\mid }z-z_0\mathrm{\mid }<\epsilon \mathrm{/}\iota (m)$ then $F(z)\in B_{[-m,m]}(F(z_0),\epsilon )$;
4. $\mathbb{R}$ is locally compact, so the exponential law applies and $\mathrm{\Phi }$ is a bijection $C(\mathbb{R}\times \mathbb{R},\mathbb{R})\to C(\mathbb{R},C(\mathbb{R},\mathbb{R}))$ whose inverse returns $f$ from $F$.

Claim 3 is the content of If $f:X\times Z\to Y$ is continuous then its transpose $F:Z\to C(X,Y)$, $F(z)(x)=f(x,z)$, is continuous for the compact-open topology, with no hypothesis on $X$ beyond being metric in this instance, verified without the tube lemma; claim 4 is where If $X$ is a locally compact metric space then the evaluation map is continuous for the compact-open topology is spent.

• Claim 3 is proved twice on purpose. The general theorem If $f:X\times Z\to Y$ is continuous then its transpose $F:Z\to C(X,Y)$, $F(z)(x)=f(x,z)$, is continuous for the compact-open topology, with no hypothesis on $X$ beyond being metric derives it from the tube lemma, and the computation above derives it from a single estimate $\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z-z_0\mathrm{\mid }\le \iota (m)\mathrm{\mid }z-z_0\mathrm{\mid }$ on $[-m,m]$. The second route is available here only because the compact sets of $\mathbb{R}$ are contained in intervals on which the first variable is bounded; the tube lemma is what replaces that boundedness in general.

• What is not checked here, because it is not claimed. Nothing above asserts that the bijection $\mathrm{\Phi }$ is a homeomorphism for any topology on the two sides. The exponential law: for a locally compact metric $X$ and any spaces $Z$ and $Y$, transposition is a bijection between $C(X\times Z,Y)$ and $C(Z,C(X,Y))$ with the compact-open topology is a bijection of sets of continuous maps, and its own remark records what the homeomorphism form would additionally require.

• The transposed family is a line of linear maps. As $z$ runs over $\mathbb{R}$ the functions $F(z)$ sweep out the multiplications by $z$, and claim 3 says that this sweep is continuous when nearness of two such maps is measured uniformly on each bounded interval. It is not continuous for the uniform metric on all of $\mathbb{R}$: for $z\mathrm{\ne }{z}_0$ the difference $\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z-z_0\mathrm{\mid }$ exceeds $1$ once $\mathrm{\mid }x\mathrm{\mid }$ is large, so the uniform distance between $F(z)$ and $F(z_0)$ is $1$ whenever $z\mathrm{\ne }{z}_0$, which is the same phenomenon as the counterexample earlier on this page.