Week 9

Section 2.9 Week 9

Subsection Monday

This class featured a discussion of mapping spaces, what they’re good for, and what we might ask of them. Given its informal nature, I won’t record notes here.

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Subsection Wednesday

Last week we introduced adjunctions and explored several examples: free \(\dashv\) forgetful adjunctions, the currying adjunction in \(\mathsf{Set}\text{,}\) and the discrete \(\dashv\) forgetful \(\dashv\) concrete adjunction for \(\mathsf{Top}\to\mathsf{Set}\text{.}\) Our goal this week is to pursue the topological currying adjunction: equipping the set of continuous maps \(\mathsf{Top}(X,Y)\) with a topology so that product and mapping space form an adjoint pair.

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Subsubsection Motivation: topologies on function spaces

For topological spaces \(X\) and \(Y\text{,}\) the set \(\mathsf{Top}(X,Y)\) of continuous maps from \(X\) to \(Y\) is just a set. Can we promote it to a topological space? And if so, which topology should we choose?

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Recall the currying adjunction in \(\mathsf{Set}\text{:}\) for a fixed set \(S\text{,}\) the functors \(S\times(-)\) and \((-)^S = \mathsf{Set}(S,-)\) satisfy

\begin{equation*} \mathsf{Set}(S\times T, U) \cong \mathsf{Set}(T, U^S). \end{equation*}

We would like an analogous statement in \(\mathsf{Top}\text{:}\) for a fixed space \(X\text{,}\) we want a topology on \(\mathsf{Top}(X,Y)\) — call the resulting space \(Y^X\) — so that

\begin{equation*} \mathsf{Top}(X\times Z, Y) \cong \mathsf{Top}(Z, Y^X) \end{equation*}

for all spaces \(Y\) and \(Z\text{.}\) When such a topology exists, we call it an exponential topology on \(\mathsf{Top}(X,Y)\text{.}\)

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Let us unpack what this adjunction would mean concretely. Reading from left to right, a continuous map \(g\colon X\times Z\to Y\) of two variables should curry to a continuous map \(\hat{g}\colon Z\to Y^X\text{;}\) that is, continuously parametrized families of continuous maps \(X\to Y\) should correspond to continuous maps into \(Y^X\text{.}\) Reading from right to left, a continuous map \(F\colon Z\to Y^X\) should uncurry to a continuous map \(X\times Z\to Y\text{;}\) the most basic instance of this (taking \(Z = Y^X\) and \(F = \mathrm{id}\)) produces the evaluation map

\begin{equation*} \mathrm{ev}\colon X\times Y^X \longrightarrow Y, \qquad (x,f)\longmapsto f(x), \end{equation*}

which should be continuous.

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Subsubsection Splitting and conjoining topologies

The discussion above suggests two properties a topology \(\tau\) on \(\mathsf{Top}(X,Y)\) might have.

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Definition 2.9.1. Splitting and conjoining topologies.

Let \(X\) and \(Y\) be topological spaces, and let \(\tau\) be a topology on \(\mathsf{Top}(X,Y)\text{.}\)

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\(\tau\) is splitting :iff, for every space \(Z\text{,}\) the continuity of \(g\colon X\times Z\to Y\) implies the continuity of the curried map \(\hat{g}\colon Z\to (\mathsf{Top}(X,Y),\tau)\text{.}\)
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\(\tau\) is conjoining :iff, for every space \(Z\text{,}\) the continuity of \(\hat{g}\colon Z\to (\mathsf{Top}(X,Y),\tau)\) implies the continuity of the uncurried map \(g\colon X\times Z\to Y\text{.}\)
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A topology that is both splitting and conjoining is an exponential topology.

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Lemma 2.9.2.

A topology \(\tau\) on \(\mathsf{Top}(X,Y)\) is conjoining if and only if the evaluation map \(\mathrm{ev}\colon X\times (\mathsf{Top}(X,Y),\tau)\to Y\) is continuous.

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Proof.

Suppose \(\mathrm{ev}\) is continuous. Given a continuous \(\hat{g}\colon Z\to (\mathsf{Top}(X,Y),\tau)\text{,}\) the uncurried map \(g = \mathrm{ev}\circ(\mathrm{id}_X\times\hat{g})\) is a composition of continuous maps, hence continuous.

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Conversely, suppose \(\tau\) is conjoining. The identity map \(\mathrm{id}\colon (\mathsf{Top}(X,Y),\tau)\to (\mathsf{Top}(X,Y),\tau)\) is continuous, so the uncurried map — which is precisely \(\mathrm{ev}\) — is continuous.

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Lemma 2.9.3.

Every splitting topology on \(\mathsf{Top}(X,Y)\) is coarser than every conjoining topology.

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Proof.

Let \(\tau\) be splitting and \(\tau'\) be conjoining. Since \(\tau'\) is conjoining, the evaluation map \(\mathrm{ev}\colon X\times(\mathsf{Top}(X,Y),\tau')\to Y\) is continuous. Since \(\tau\) is splitting, the curried map \(\mathrm{id}\colon (\mathsf{Top}(X,Y),\tau')\to (\mathsf{Top}(X,Y),\tau)\) is continuous, which means \(\tau\subseteq\tau'\text{.}\)

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Theorem 2.9.4. Uniqueness of exponential topologies.

If an exponential topology on \(\mathsf{Top}(X,Y)\) exists, then it is unique.

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Proof.

An exponential topology is both splitting and conjoining, so by Lemma 2.9.3, any two exponential topologies are each coarser than the other.

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Subsubsection The compact-open topology

We now define the compact-open topology and show it is always splitting. The key question — when is it also conjoining? — will be taken up on Friday.

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Definition 2.9.5. Compact-open topology.

Let \(X\) and \(Y\) be topological spaces. For each compact subspace \(K\subseteq X\) and each open subset \(U\subseteq Y\text{,}\) define

\begin{equation*} [K,U] := \{f\in\mathsf{Top}(X,Y)\mid f(K)\subseteq U\}. \end{equation*}

The sets \([K,U]\) form a subbasis for a topology on \(\mathsf{Top}(X,Y)\) called the compact-open topology. We write \(Y^X\) for \(\mathsf{Top}(X,Y)\) equipped with this topology.

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Remark 2.9.6.

The subbasis sets \([K,U]\) form a subbasis, not generally a basis. Their finite intersections

\begin{equation*} [K_1,U_1]\cap\cdots\cap [K_n,U_n] \end{equation*}

form a basis.

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Example 2.9.7.

Let \(X=\{1,\ldots,r\}\) with the discrete topology and let \(Y\) be any topological space. Then \(\mathsf{Top}(X,Y) = Y^r\) as a set. The compact subsets of \(X\) are all subsets (since \(X\) is finite), and the subbasis sets \([\{k\},U] = \{(y_1,\ldots,y_r)\mid y_k\in U\}\) generate the product topology on \(Y^r\text{.}\) So in this case the compact-open topology agrees with the (finite) product topology.

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This observation will guide us toward the correct definition of the product topology for arbitrary (possibly infinite) families of spaces.

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Theorem 2.9.8. The compact-open topology is splitting.

For any spaces \(X\) and \(Y\text{,}\) the compact-open topology on \(\mathsf{Top}(X,Y)\) is splitting.

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Proof.

Let \(Z\) be any space and suppose \(g\colon X\times Z\to Y\) is continuous. We need to show that the curried map \(\hat{g}\colon Z\to Y^X\) is continuous. It suffices to show that \(\hat{g}^{-1}([K,U])\) is open in \(Z\) for each compact \(K\subseteq X\) and open \(U\subseteq Y\text{.}\) Now, \(\hat{g}^{-1}([K,U]) = \{z\in Z\mid g(K,z)\subseteq U\}\text{.}\) Since \(g\) is continuous, \(g^{-1}(U)\) is open in \(X\times Z\) and contains \(K\times\{z\}\text{.}\) By the Tube Lemma, there are open sets \(K\subseteq V\) and \(z\in W\) with \(K\times\{z\}\subseteq V\times W\subseteq g^{-1}(U)\text{.}\) Therefore \(z\in W\subseteq \hat{g}^{-1}([K,U])\text{.}\)

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Subsubsection Metrics on function spaces

When the target space \((Y,d)\) is a metric space, one can define several natural topologies on the set \(\mathsf{Top}(X,Y)\) via metrics or pseudometrics. For a subset \(A\subseteq X\text{,}\) a map \(f\colon X\to Y\text{,}\) and \(\varepsilon\gt 0\text{,}\) write

\begin{equation*} \mathrm{B}_\varepsilon^A(f) := \{g\in\mathsf{Top}(X,Y)\mid \sup_{x\in A}d(f(x),g(x))\lt\varepsilon\} \end{equation*}

for the \(\varepsilon\)-ball about \(f\) measured on \(A\text{.}\) Varying \(A\) yields three topologies on \(\mathsf{Top}(X,Y)\text{.}\)

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Pointwise convergence. Take \(A = \{x\}\) for each \(x\in X\text{.}\) The resulting topology \(\tau_{\mathrm{pw}}\) on \(\mathsf{Top}(X,Y)\) coincides with the subspace topology from the product \(Y^X = \prod_{x\in X}Y\text{.}\) A sequence \((f_n)\) converges to \(f\) in this topology if and only if \(f_n(x)\to f(x)\) for each \(x\in X\text{.}\)
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Compact convergence. Take \(A = K\) for each compact \(K\subseteq X\text{.}\) The resulting topology \(\tau_{\mathrm{cpt}}\) is coarser than the uniform topology and finer than the pointwise topology. A sequence \((f_n)\) converges to \(f\) if and only if \((f_n|_K)\) converges uniformly to \(f|_K\) for every compact \(K\subseteq X\text{.}\) When \(Y\) is a metric space, the compact convergence topology on \(\mathsf{Top}(X,Y)\) agrees with the compact-open topology.
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Uniform convergence. Take \(A = X\text{.}\) When \(X\) is compact, this agrees with the compact convergence topology (since \(X\) itself is compact). In general, the uniform topology is finer: it requires convergence to be uniform over all of \(X\) simultaneously. When \(X\) is compact, the sup metric \(d_\infty(f,g) := \sup_{x\in X}d(f(x),g(x))\) defines the topology, and \((\mathsf{Top}(X,Y),d_\infty)\) is a metric space.
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Summarizing: there is a chain of inclusions among topologies on \(\mathsf{Top}(X,Y)\text{,}\)

\begin{equation*} \tau_{\mathrm{pw}}\subseteq\tau_{\mathrm{cpt}}\subseteq\tau_{\mathrm{unif}}, \end{equation*}

and when \(X\) is compact, the latter two coincide. We omit the (straightforward but somewhat technical) proofs that the compact convergence and compact-open topologies agree; see Section 11.5 of the graduate notes for details. An important consequence of this agreement is that a classical theorem — the uniform limit of continuous functions is continuous — can be reinterpreted as saying that \(\mathsf{Top}(X,Y)\) is a closed subspace of the space of all maps \(X\to Y\) in the compact convergence topology.

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Subsection Friday

Subsubsection The product topology via mapping spaces

So far in this course, we have only defined products of finitely many topological spaces: given spaces \(X_1,\ldots,X_n\text{,}\) the product \(X_1\times\cdots\times X_n\) carries the topology generated by products of open sets \(U_1\times\cdots\times U_n\text{.}\) The compact-open topology gives us a conceptual way to extend this definition to arbitrary (possibly infinite) products.

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The key observation is Example 2.9.7: when \(A\) is a finite discrete space, the compact-open topology on \(Y^A = \mathsf{Top}(A,Y)\) recovers the product topology on the finite product \(Y^{|A|}\text{.}\) But the definition of the compact-open topology makes perfect sense when \(A\) is an infinite discrete set — and this tells us what the right topology on an infinite product should be.

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Let \(A\) be any set, equipped with the discrete topology. Since every function out of a discrete space is continuous, we have \(\mathsf{Top}(A,Y) = Y^A\) as sets. The compact-open subbasis sets are \([K,U]\) for \(K\subseteq A\) compact and \(U\subseteq Y\) open. Since \(A\) is discrete, its compact subsets are precisely its finite subsets. The subbasis set \([\{\alpha\},U]\) for a single element \(\alpha\in A\) is

\begin{equation*} [\{\alpha\},U] = \{f\in Y^A\mid f(\alpha)\in U\} = \mathrm{pr}_\alpha^{-1}(U), \end{equation*}

where \(\mathrm{pr}_\alpha\colon Y^A\to Y\) is projection onto the \(\alpha\)-th coordinate. In other words, the compact-open topology on \(Y^A\) is the coarsest topology making every projection continuous. We promote this observation to a definition.

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Definition 2.9.9. Product topology.

Let \(\{Y_\alpha\}_{\alpha\in A}\) be a family of topological spaces. The product topology on the set \(\prod_{\alpha\in A}Y_\alpha\) is the coarsest topology for which every projection map

\begin{equation*} \mathrm{pr}_\alpha\colon \prod_{\beta\in A}Y_\beta\longrightarrow Y_\alpha, \qquad (y_\beta)_{\beta\in A}\longmapsto y_\alpha, \end{equation*}

is continuous.

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Equivalently, the product topology has subbasis

\begin{equation*} \bigl\{\mathrm{pr}_\alpha^{-1}(U)\mid \alpha\in A,\;U\subseteq Y_\alpha\text{ open}\bigr\}. \end{equation*}

A basis consists of all finite intersections of these subbasis sets, i.e., sets of the form

\begin{equation*} \mathrm{pr}_{\alpha_1}^{-1}(U_1)\cap\cdots\cap\mathrm{pr}_{\alpha_n}^{-1}(U_n) \end{equation*}

where \(\alpha_1,\ldots,\alpha_n\in A\) and each \(U_i\subseteq Y_{\alpha_i}\) is open.

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Remark 2.9.10.

When \(Y_\alpha = Y\) for all \(\alpha\text{,}\) the product \(\prod_\alpha Y\) is the set \(Y^A\) of functions \(A\to Y\text{.}\) If \(A\) carries the discrete topology, this is \(\mathsf{Top}(A,Y)\text{,}\) and the product topology is exactly the compact-open topology, as argued above. For a general (non-discrete) space \(X\text{,}\) the product topology on \(Y^X\) has subbasis sets \([\{x\},U] = \mathrm{pr}_x^{-1}(U)\) indexed by singletons, while the compact-open topology allows arbitrary compact \(K\text{.}\) Since singletons are compact, the compact-open topology is always finer than the product topology.

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As always, we should ask: what is the universal property? The product topology is designed exactly so that continuous maps into a product are determined by their coordinates.

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Theorem 2.9.11. Universal property of products.

Let \(\{Y_\alpha\}_{\alpha\in A}\) be a family of topological spaces. For any topological space \(Z\text{,}\) a map

\begin{equation*} f\colon Z\longrightarrow \prod_{\alpha\in A}Y_\alpha \end{equation*}

is continuous if and only if \(\mathrm{pr}_\alpha\circ f\colon Z\to Y_\alpha\) is continuous for every \(\alpha\in A\text{.}\)

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In other words, there is a bijection of sets

\begin{equation*} \mathsf{Top}\!\left(Z,\,\prod_{\alpha\in A}Y_\alpha\right) \xrightarrow{\;\cong\;} \prod_{\alpha\in A}\mathsf{Top}(Z,Y_\alpha), \qquad f\longmapsto (\mathrm{pr}_\alpha\circ f)_{\alpha\in A}, \end{equation*}

natural in \(Z\text{.}\)

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Proof.

The "only if" direction is immediate: if \(f\) is continuous, then \(\mathrm{pr}_\alpha\circ f\) is a composition of continuous maps, hence continuous.

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For the "if" direction, suppose \(\mathrm{pr}_\alpha\circ f\) is continuous for every \(\alpha\text{.}\) To show \(f\) is continuous, it suffices to check preimages of subbasis sets. For each subbasis set \(\mathrm{pr}_\alpha^{-1}(U)\) in the product, we have

\begin{equation*} f^{-1}\bigl(\mathrm{pr}_\alpha^{-1}(U)\bigr) = (\mathrm{pr}_\alpha\circ f)^{-1}(U), \end{equation*}

which is open in \(Z\) by the assumed continuity of \(\mathrm{pr}_\alpha\circ f\text{.}\)

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Remark 2.9.12.

The product topology is the unique topology on \(\prod_\alpha Y_\alpha\) satisfying this universal property. Indeed, if \(\tau\) is any topology making the projections continuous and satisfying the universal property, then the identity map \(\mathrm{id}\colon(\prod_\alpha Y_\alpha,\tau_{\mathrm{prod}})\to(\prod_\alpha Y_\alpha,\tau)\) is continuous (since each projection from the product topology is continuous), and vice versa.

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We record an important special case connecting back to our discussion of mapping spaces.

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Corollary 2.9.13.

Let \(A\) be a set (with discrete topology) and \(Y\) a topological space. For any space \(Z\text{,}\) there is a natural bijection

\begin{equation*} \mathsf{Top}(Z,Y^A)\xrightarrow{\;\cong\;}\mathsf{Top}(Z,Y)^A, \end{equation*}

where \(Y^A = \prod_{\alpha\in A}Y\) carries the product topology. That is, a continuous map from \(Z\) into \(Y^A\) is the same thing as an \(A\)-indexed family of continuous maps \(Z\to Y\text{.}\)

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Subsubsection Local compactness

On Wednesday we showed that the compact-open topology is always splitting. The question remains: when is it conjoining? The answer hinges on a topological property called local compactness. Before stating the main theorem, let us introduce this condition and develop some intuition for it.

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Definition 2.9.14. Locally compact space.

A topological space \(X\) is locally compact if for every point \(x\in X\) and every open neighborhood \(U\) of \(x\text{,}\) there exists a compact subspace \(K\subseteq X\) such that

\begin{equation*} x\in\mathrm{Int}(K)\subseteq K\subseteq U. \end{equation*}

In other words, every point has a neighborhood basis of compact sets.

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Remark 2.9.15.

Some authors use a weaker definition, requiring only that every point has some compact neighborhood. For Hausdorff spaces the two definitions are equivalent, and all of our applications will be in the Hausdorff setting.

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Example 2.9.16. Locally compact spaces.

Euclidean spaces. \(\mathbb{R}^n\) is locally compact: given \(x\in U\subseteq\mathbb{R}^n\) with \(U\) open, choose \(\varepsilon\gt 0\) so that the open ball \(B_\varepsilon(x)\subseteq U\text{,}\) and take \(K\) to be the closed ball \(\overline{B}_{\varepsilon/2}(x)\text{,}\) which is compact by the Heine-Borel theorem.
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Compact spaces. Every compact space is locally compact: take \(K = X\text{.}\)
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Discrete spaces. Every discrete space is locally compact: the singleton \(\{x\}\) is a compact (indeed finite) neighborhood of \(x\text{.}\)
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Open subsets of locally compact Hausdorff spaces. If \(X\) is locally compact Hausdorff and \(V\subseteq X\) is open, then \(V\) is locally compact in the subspace topology.
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Manifolds. Every topological manifold is locally compact, since each point has a neighborhood homeomorphic to an open subset of \(\mathbb{R}^n\text{.}\)
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Example 2.9.17. Spaces that are not locally compact.

The rational numbers \(\mathbb{Q}\) with the subspace topology inherited from \(\mathbb{R}\) are not locally compact. Indeed, a compact subset of \(\mathbb{Q}\) has empty interior in \(\mathbb{Q}\) (since every open interval in \(\mathbb{Q}\) is homeomorphic to \(\mathbb{Q}\text{,}\) which is not compact), so no point has a compact neighborhood with nonempty interior.
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Infinite-dimensional spaces. The space \(\mathbb{R}^\infty = \bigcup_{n\ge 0}\mathbb{R}^n\) (with the colimit topology) is not locally compact. Similarly, the Hilbert space \(\ell^2\) is not locally compact: the closed unit ball is not compact, and more generally no point has a compact neighborhood.
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With local compactness in hand, we can now state and prove the key result: the compact-open topology is conjoining — and therefore exponential — whenever the domain is locally compact.

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Subsubsection Continuity of evaluation and the exponential adjunction

By Lemma 2.9.2, the compact-open topology is conjoining if and only if the evaluation map \(\mathrm{ev}\colon X\times Y^X\to Y\) is continuous. We now show that local compactness of the domain is exactly the right condition.

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Lemma 2.9.18. Continuity of the evaluation map.

Let \(X\) and \(Y\) be topological spaces. If \(X\) is locally compact, then the evaluation map

\begin{equation*} \mathrm{ev}\colon X\times Y^X \longrightarrow Y, \qquad (x,f)\longmapsto f(x), \end{equation*}

is continuous, where \(Y^X = \mathsf{Top}(X,Y)\) carries the compact-open topology.

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Proof.

Let \(U\subseteq Y\) be open. We must show the preimage \(\mathrm{ev}^{-1}(U)\subseteq X\times Y^X\) is open. Let \((x,f)\in\mathrm{ev}^{-1}(U)\text{,}\) so \(f(x)\in U\text{.}\) Since \(f\) is continuous, \(f^{-1}(U)\) is an open neighborhood of \(x\) in \(X\text{.}\) Since \(X\) is locally compact, there is a compact subspace \(K\subseteq X\) with

\begin{equation*} x\in\mathrm{Int}(K)\subseteq K\subseteq f^{-1}(U). \end{equation*}

Then \(f(K)\subseteq U\text{,}\) so \(f\in [K,U]\text{.}\) The set \(\mathrm{Int}(K)\times [K,U]\) is an open neighborhood of \((x,f)\) in \(X\times Y^X\text{.}\) For any \((x',f')\) in this neighborhood, we have \(x'\in\mathrm{Int}(K)\subseteq K\) and \(f'(K)\subseteq U\text{,}\) so \(f'(x')\in U\text{,}\) confirming \(\mathrm{Int}(K)\times [K,U]\subseteq\mathrm{ev}^{-1}(U)\text{.}\)

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Theorem 2.9.19. The exponential adjunction for locally compact spaces.

Let \(X\) be a locally compact topological space and let \(Y\) be any topological space. Then the compact-open topology on \(\mathsf{Top}(X,Y)\) is exponential. That is, for every space \(Z\text{,}\) currying and uncurrying define inverse bijections

\begin{equation*} \mathsf{Top}(X\times Z, Y) \cong \mathsf{Top}(Z, Y^X) \end{equation*}

that are natural in \(Y\) and \(Z\text{.}\) In the language of adjunctions, this says

\begin{equation*} X\times(-)\;\dashv\;(-)^X\colon\mathsf{Top}\longrightarrow\mathsf{Top}. \end{equation*}

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Proof.

By Theorem 2.9.8, the compact-open topology is splitting, and by Lemma 2.9.18 and Lemma 2.9.2, it is conjoining when \(X\) is locally compact.

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Subsubsection Applications of the exponential adjunction

The exponential adjunction has many pleasant consequences. We record several here.

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Corollary 2.9.20. Continuity of composition.

Let \(X\text{,}\) \(Y\text{,}\) and \(Z\) be topological spaces with \(X\) and \(Y\) locally compact. Then the composition map

\begin{equation*} \circ\colon Y^X\times Z^Y\longrightarrow Z^X, \qquad (f,g)\longmapsto g\circ f, \end{equation*}

is continuous.

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Proof.

The uncurrying of the composition map is the continuous map

\begin{equation*} X\times Y^X\times Z^Y \xrightarrow{\;\mathrm{ev}\times\mathrm{id}\;} Y\times Z^Y \xrightarrow{\;\mathrm{ev}\;} Z, \qquad (x,f,g)\longmapsto g(f(x)). \end{equation*}

Both evaluation maps are continuous (since \(X\) and \(Y\) are locally compact), as are products and compositions of continuous maps.

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Corollary 2.9.21. Exponential law for mapping spaces.

Let \(X\) and \(Z\) be locally compact and let \(Y\) be any space. The currying bijection

\begin{equation*} Y^{X\times Z}\xrightarrow{\;\cong\;} (Y^X)^Z \end{equation*}

is a homeomorphism. This categorifies the identity \(y^{xz}=(y^x)^z\text{.}\)

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Proof.

This is an exercise. (Hint: one can give a clean proof using the fact that both sides represent the same functor, appealing to the Yoneda lemma.)

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Subsubsection Limitations and the role of local compactness

It is natural to ask how essential the hypothesis of local compactness is. The short answer: very.

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The category \(\mathsf{Top}\) is not Cartesian closed: there is no way to functorially equip \(\mathsf{Top}(X,Y)\) with a topology making the product-hom adjunction work for all spaces \(X\text{.}\) Locally compact Hausdorff spaces are well-behaved in this respect, but problems arise even for simple spaces.

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Example 2.9.22.

Consider the colimit of the diagram

\begin{equation*} \mathbb{R}\hookrightarrow\mathbb{R}^2\hookrightarrow\mathbb{R}^3\hookrightarrow\cdots \end{equation*}

in \(\mathsf{Top}\text{.}\) The resulting space \(\mathbb{R}^\infty\) is not locally compact, so the exponential adjunction may fail for mapping spaces with domain \(\mathbb{R}^\infty\text{.}\)

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One remedy is to restrict attention to a convenient category of topological spaces — one that is Cartesian closed and contains all the spaces we care about. The category of compactly generated weakly Hausdorff (CGWH) spaces is the standard choice. A topological space \(X\) is compactly generated if \(A\subseteq X\) is open whenever \(A\cap K\) is open in \(K\) for every compact \(K\subseteq X\text{.}\) Locally compact spaces are compactly generated, and the CGWH category has enough structure to support mapping spaces, quotients, and all the constructions we need for algebraic topology.

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We will not develop the theory of CGWH spaces in this course, but it is good to be aware that the exponential adjunction can be rescued by working in an appropriate subcategory. In practice, all the spaces we encounter (manifolds, CW complexes, metric spaces, etc.) are locally compact Hausdorff, and the compact-open topology works beautifully for them.

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Subsubsection Application: Stone-Čech compactification via mapping spaces

Recall from Subsubsection that for any topological space \(X\text{,}\) the Stone-Čech compactification \(\beta X\) is a compact Hausdorff space equipped with a continuous map \(\eta_X\colon X\to\beta X\) such that every continuous map from \(X\) to a compact Hausdorff space \(Y\) extends uniquely across \(\eta_X\text{.}\) With the compact-open topology in hand, we can at least describe how one constructs \(\beta X\text{.}\)

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Consider the unit interval \([0,1]\text{,}\) which is compact Hausdorff. For any completely regular space \(X\text{,}\) the evaluation map

\begin{equation*} X\longrightarrow [0,1]^{\mathsf{Top}(X,[0,1])}, \qquad x\longmapsto (\mathrm{ev}_x\colon f\mapsto f(x)), \end{equation*}

is an embedding. The closure of the image is a compact Hausdorff space, and this is \(\beta X\text{.}\)

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