nLab maps from compact spaces to Hausdorff spaces are closed and proper

Statement

Proposition

(maps from compact spaces to Hausdorff spaces are closed and proper)

Let $f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y)$ be a continuous function between topological spaces such that

1. $(X,\tau_X)$ is a compact topological space;

2. $(Y,\tau_Y)$ is a Hausdorff topological space.

Then $f$ is

1. a closed map;

2. a proper map.

Proof

For the first statement, we need to show that if $C \subset X$ is a closed subset of $X$, then also $f(C) \subset Y$ is a closed subset of $Y$.

Now

1. since closed subsets of compact spaces are compact it follows that $C \subset X$ is also compact;

2. since continuous images of compact spaces are compact it then follows that $f(C) \subset Y$ is compact;

3. since compact subspaces of Hausdorff spaces are closed it finally follow that $f(C)$ is also closed in $Y$.

For the second statement we need to show that if $C \subset Y$ is a compact subset, then also its pre-image $f^{-1}(C)$ is compact.

Now

1. since compact subspaces of Hausdorff spaces are closed it follows that $C \subset Y$ is closed;

2. since pre-images under continuous of closed subsets are closed, also $f^{-1}(C) \subset X$ is closed;

3. since closed subsets of compact spaces are compact, it follows that $f^{-1}(C)$ is compact.

Consequences

Corollary

(continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms)

Let $f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y)$ be a continuous function between topological spaces such that

1. $(X,\tau_X)$ is a compact topological space;

2. $(Y,\tau_Y)$ is a Hausdorff topological space.

3. $f \;\colon\; X \longrightarrow Y$ is a bijection of sets.

Then $f$ is a homeomorphism, i. e. its inverse function $Y \to X$ is also a continuous function.

In particular then both $(X,\tau_X)$ and $(Y, \tau_Y)$ are compact Hausdorff spaces.

Proof

Write $g \colon Y \to X$ for the inverse function of $f$.

We need to show that $g$ is continuous, hence that for $U \subset X$ an open subset, then also its pre-image $g^{-1}(U) \subset Y$ is open in $Y$. By passage to complements, this is equivalent to the statement that for $C \subset X$ a closed subset then the pre-image $g^{-1}(C) \subset Y$ is also closed in $Y$.

But since $g$ is the inverse function to $f$, its pre-images are the images of $f$. Hence the last statement above equivalently says that $f$ sends closed subsets to closed subsets. This is true by prop. .

Remark

The idea captured by corollary is that Hausdorffness is about having “enough” open sets whilst compactness is about having “not too many”. Thus a compact Hausdorff space has both “enough” and “not too many”. This theorem says that both conditions are at their limit: if we try to have more open sets, we lose compactness. If we try to have fewer open sets, we lose Hausdorffness.

Applications

In Cohomotopy and Cobordism theory

Pontryagin's theorem establishes a bijection

$$\pi^n \big( X^d \big) \,=\, \pi_0 Maps \big( X^d, S^n \big) \overset {\simeq}{ \;\; \longrightarrow \;\; } Cob^n_{Fr} \big( X^d \big)$$

between

1. the $n$-Cohomotopy set of a closed (hence: compact) smooth manifold $X^d$;

2. the cobordism classes of its normally framed submanifolds of codimension $n$;

by taking the homotopy class of any map $[X \overset{c}{\to}S^n]$ into the n-sphere to the preimage of any regular point (say $\{0\} \subset S^n$, for definiteness) of a smooth representative $c$:

$$\Sigma_c \;\coloneqq\; c^{-1}\big( \{0\} \big) \,.$$

To show that this construction is indeed injective, one needs that, similarly, the pre-image of any regular point of a smooth homotopy

$$X^d \times [0,1] \overset{\eta}{\longrightarrow} S^d$$

between two such smooth representatives $c_0, c_1$ is a cobordism between $\Sigma_{c_0}$ and $\Sigma_{c_1}$. But for a subspace of $X^d \times [0,1]$ to constitute a cobordism between submanifolds of $X^d$ it is necessary that its projection onto the $[0,1]$-factor is compact.

That this is implied here is guaranteed by the assumption that $X^d$ is closed, hence compact, so that also the product space $X^d \times [0,1]$ is compact:

With this and since the n-sphere is, of course, Hausdorff, Prop. implies that the maps $c_i$ and $\eta$ above are all proper, hence that the corresponding pre-images (of singleton, hence compact, subspaces) are indeed compact, so that in particular the projection of the pre-image of $\eta$ to $[0,1]$ is compact (since continuous images of compact spaces are compact).

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• closed subsets of compact spaces are compact

• compact subspaces of Hausdorff spaces are closed

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