What Compactness Tells Us About an Image
The Compactness in General Topology tutorial established that a continuous image of a compact space is compact, when the image is given the subspace topology of the codomain. This result does not require the codomain to be Hausdorff. It tells us that compactness follows the map to the part of the target that is actually reached, even if the whole target has no compactness property.
That distinction is useful. A continuous map need not be onto, so the compactness of its image does not imply that its codomain is compact. When the codomain is Hausdorff, compact images have an additional consequence: they are closed. Combining these facts gives a powerful way to show that certain continuous maps are closed, and then to recognize when a continuous bijection is a homeomorphism.
The phrase “with the subspace topology” is essential. Compactness is a property of a topological space, not merely of an underlying set. In this tutorial, whenever we call \(f[X]\) compact, we mean that it has the topology inherited from \(Y\). The image can be compact even when \(Y\) itself is not.
Compact Domains Make Maps into Hausdorff Spaces Closed
A map \(f:X\to Y\) is called closed if it sends every closed subset of \(X\) to a closed subset of \(Y\). Continuity concerns inverse images of open sets; closedness concerns direct images of closed sets. These are different properties in general. Compactness of the domain supplies a useful bridge between them when the target is Hausdorff.
Proof. Let \(F\) be any closed subset of \(X\). By the theorem that closed subsets of compact spaces are compact, \(F\) is compact. The restriction of \(f\) to \(F\) is continuous, so the Compact-Image Theorem shows that \(f[F]\) is compact as a subspace of \(Y\). Since \(Y\) is Hausdorff, the theorem that compact subsets of Hausdorff spaces are closed applies: \(f[F]\) is closed in \(Y\). This holds for every closed \(F\subseteq X\), so \(f\) is a closed map. \(\square\)
This proof uses two results from “Compactness in General Topology”: closed subsets of compact spaces are compact, and compact subsets of Hausdorff spaces are closed. The roles of the hypotheses should be kept distinct. Compactness of \(X\) first gives compactness of each closed \(F\); continuity transfers that compactness to \(f[F]\); Hausdorffness of \(Y\) then makes the image closed.
Worked Example: A Projection Sends Closed Sets to Closed Sets
Let \(X=[0,1]\times[0,1]\), with its usual product topology, and let \(p:X\to[0,1]\) be the first-coordinate projection, \(p(x,y)=x\). The square is compact, and \([0,1]\) is Hausdorff. The projection is continuous, so the theorem says that \(p\) sends every closed subset of the square to a closed subset of \([0,1]\).
For a specific instance, take $$ F=\{(x,y)\in[0,1]\times[0,1]:x+y\leq 1\}. $$ The function \(s(x,y)=x+y\) is continuous, and \((-\infty,1]\) is closed in \(\mathbb R\), so \(F=s^{-1}((-\infty,1])\) is closed in the square. Its projection is $$ p[F]=[0,1]. $$ Indeed, if \((x,y)\in F\), then \(x\in[0,1]\), so \(p[F]\subseteq[0,1]\). Conversely, for every \(x\in[0,1]\), the point \((x,0)\) belongs to \(F\), because \(x+0\leq1\). Thus \(x=p(x,0)\in p[F]\), proving \([0,1]\subseteq p[F]\). The example illustrates the conclusion for one closed set; the theorem guarantees it for every closed subset of the square.
Continuous Bijections and Homeomorphisms
A continuous bijection has a continuous inverse exactly when it is a homeomorphism. Continuity of the forward map alone is not usually enough. The closed-map theorem gives a convenient way to verify continuity of the inverse: for a bijection \(f:X\to Y\), the inverse image under \(f^{-1}\) of a set \(F\subseteq X\) is \(f[F]\). If \(f\) sends closed sets to closed sets, those inverse images are closed, so the closed-set characterization of continuity applies to \(f^{-1}\).
Proof. The closed-map theorem shows that \(f\) is closed. We prove that its inverse \(f^{-1}:Y\to X\) is continuous. Let \(F\) be closed in \(X\). Then $$ (f^{-1})^{-1}[F]=f[F]. $$ Because \(f\) is closed, \(f[F]\) is closed in \(Y\). Thus the inverse image under \(f^{-1}\) of every closed subset of \(X\) is closed in \(Y\). By the closed-set characterization of continuity, \(f^{-1}\) is continuous. Since \(f\) is already continuous and bijective, it is a homeomorphism. \(\square\)
In practice, this theorem often avoids a direct attempt to prove that the inverse is continuous. One verifies compactness of the domain, Hausdorffness of the target, and continuity and bijectivity of the map. The topology-preserving property of the inverse then follows.
Worked Example: A Continuous Bijection from an Interval onto a Circle
Define \(f:[0,2\pi]\to\mathbb R^2\) by $$ f(t)=(\cos t,\sin t). $$ The coordinate functions are continuous, so \(f\) is continuous. Its image lies on the unit circle \(S^1=\{(x,y)\in\mathbb R^2:x^2+y^2=1\}\), because for each \(t\), $$ (\cos t)^2+(\sin t)^2=1. $$ Every point of \(S^1\) has the form \((\cos t,\sin t)\) for some \(t\in[0,2\pi]\), so \(f\) is onto \(S^1\). The domain interval is compact, and \(S^1\), as a subspace of the Hausdorff space \(\mathbb R^2\), is Hausdorff.
However, \(f\) is not injective: direct substitution gives \(f(0)=(1,0)\) and \(f(2\pi)=(1,0)\), although \(0\ne2\pi\). Thus the compact-to-Hausdorff theorem does not say that every continuous map from a compact space to a Hausdorff space is a homeomorphism. Bijectivity is a necessary hypothesis in the theorem. The Compact-Image Theorem still guarantees that the image \(S^1\) is compact.
Why Hausdorffness Matters
The Hausdorff condition cannot be dropped from the continuous-bijection theorem. If the target is not Hausdorff, a continuous bijection from a compact space need not have a continuous inverse. The following example uses familiar topologies to show exactly where the conclusion can fail.
Worked Example: A Compact Domain Does Not Suffice for a Homeomorphism
Let \(S\) be an infinite set. Give \(X=S\) the cofinite topology: its open sets are \(\varnothing\) and the subsets whose complements in \(S\) are finite. Give \(Y=S\) the indiscrete topology, whose only open sets are \(\varnothing\) and \(S\). The cofinite topology on an infinite set is compact, as established earlier in the course.
Consider the identity map \(f:X\to Y\). It is bijective. It is also continuous: the inverse image of either open set in \(Y\) is either \(\varnothing\) or \(S\), both of which are open in \(X\). But \(f^{-1}:Y\to X\) is not continuous. Choose \(s\in S\). The set \(S\setminus\{s\}\) is open in \(X\), since its complement is finite and \(S\setminus\{s\}\) is nonempty. Its inverse image under \(f^{-1}\) is the same set \(S\setminus\{s\}\), which is not open in the indiscrete space \(Y\). Therefore \(f^{-1}\) is not continuous, and \(f\) is not a homeomorphism. This does not contradict the theorem: \(Y\) is not Hausdorff.
There is another useful distinction: a compact image need not be closed if the codomain is not Hausdorff. The theorem that compact subsets of Hausdorff spaces are closed has a genuine separation hypothesis. A compact subset of a non-Hausdorff space may fail to be closed, so the argument turning a compact image into a closed image cannot proceed without that hypothesis.
Using the Results Carefully
The Compact-Image Theorem, the closed-map theorem, and the continuous-bijection theorem answer different questions. The first concerns \(f[X]\) and needs only a continuous map from a compact domain. The second concerns images of all closed subsets of \(X\) and also requires a Hausdorff codomain. The third adds bijectivity and concludes that the inverse is continuous.
| Claim | Additional hypotheses | Conclusion |
|---|---|---|
| A continuous map carries a compact domain to a compact image. | Continuity and compactness of the domain. | \(f[X]\) is compact in its subspace topology. |
| A continuous map from a compact space is closed. | The codomain is Hausdorff. | Images of closed subsets are closed. |
| A continuous bijection is a homeomorphism. | The domain is compact and the codomain is Hausdorff. | The inverse map is continuous. |
A common error is to conclude that the codomain is compact whenever the domain is compact and the map is continuous. The conclusion applies to the image, not necessarily to all of \(Y\). For example, a constant map from \([0,1]\) into \(\mathbb R\) has a one-point image, which is compact, while its codomain \(\mathbb R\) is not compact. Surjectivity would make the image equal to \(Y\), but without surjectivity the two sets must not be confused.
When applying the homeomorphism theorem, check all of its hypotheses rather than just continuity and compactness. A map that is not bijective cannot have an inverse function on the whole codomain, as the circle parametrization demonstrates. A bijection whose target is not Hausdorff may have a discontinuous inverse, as the cofinite-to-indiscrete example demonstrates. The hypotheses work together to turn compactness into control over the topology of the image.
Check Your Understanding
Use the Compact-Image Theorem and its consequences to answer the following questions.
- When a continuous map has compact domain, which space is guaranteed to be compact: the codomain or the image with its subspace topology?
- In the closed-map theorem, where is compactness used, and where is Hausdorffness used?
- Why does a continuous bijection from a compact space to a Hausdorff space have a continuous inverse?
- For the map \(t\mapsto(\cos t,\sin t)\) on \([0,2\pi]\), verify why the image lies on the unit circle and identify the failure of injectivity.
- In the cofinite-to-indiscrete example, which set is open in the cofinite topology but not in the indiscrete topology, and what does that show about the inverse map?