Home > Informal Classroom Notes > Why does Compactness ensure a Bijective Map is a Homeomorphism?

## Why does Compactness ensure a Bijective Map is a Homeomorphism?

The motivation for this exposition is threefold:

• It helps illustrate the mantra that intuition is as equally important as algebraic manipulation.
• It relates to a fundamental concept in mathematics, namely morphisms (from category theory).
• It helps explain why compact spaces and Hausdorff spaces are important.

The following facts are readily found in textbooks; the tenet of this exposition is that it is impossible, without further thought, to appreciate any of these facts.

A topological space is compact if every open cover has a finite subcover.

A topological space is Hausdorff if every two points can be separated by neighbourhoods.

A map (i.e., a continuous function) $f \colon X \rightarrow Y$ between topological spaces $X$ and $Y$ is a homeomorphism if it is bijective and its inverse $f^{-1}$ is continuous.

If $X$ is compact and $Y$ is Hausdorff then a bijective map $f \colon X \rightarrow Y$ is automatically a homeomorphism.

Questions:

• How can a bijective map not be a homeomorphism?
• How can compactness possibly have any relevance to whether a map is a homeomorphism?
• Why the Hausdorff assumption on $Y$?
• How does this relate to “morphisms”, as mentioned at the start of this exposition?

Before addressing these questions, a standard proof of the above result is given, the purpose being to emphasise that the proof, on its own, does not directly answer any of these questions. Memorising the proof is essentially worthless. (Important are being able to derive the proof from scratch and being able to understand why the result should indeed be true.)

Showing $f^{-1}$ is continuous means showing the preimage of every open set is open, or equivalently, showing the preimage of every closed set is closed. Since $f$ is bijective, the preimage under $f^{-1}$ of a set $A$ is simply $f(A)$. Hence it suffices to prove that $f$ is closed (the image of every closed set is closed). Let $A \subset X$ be closed. Since $X$ is compact, $A$ must be compact. The image of a compact set under a continuous function is itself compact, that is, $f(A)$ is compact. A compact subset of a Hausdorff space must be closed, that is, $f(A)$ is a closed subset of $Y$, proving $f$ is closed and therefore that $f^{-1}$ is continuous, as required.

While basing topology on open sets works very well from the perspective of being able to give concise proofs such as the one above, it is perhaps the worst way of introducing topology. It is noteworthy that when topology was being developed, it took decades for the importance of “open sets” to be recognised.

Let’s start from scratch and try to understand things intuitively. One approach (but certainly not the only approach) to understanding a topological space is understanding what converges to what. In special cases, such as when the topology comes from a metric, it suffices to consider sequences (the curious reader may wish to read about sequential spaces). In general, nets must be considered instead of sequences, nevertheless, for the purposes of answering the questions above, it suffices to restrict attention to sequences. (It is often enough to understand intuitively why a result is true in special cases, leaving the proof to justify the result in full generality.)

## Why a Bijective Map need not be a Homeomorphism

By restricting attention to sequences (and therefore to nice topological spaces), the definition of continuity becomes: $f$ is continuous if, whenever $x_n \rightarrow x$, it holds that $f(x_n) \rightarrow f(x)$. Therefore, a bijective map $f$ is a homeomorphism if and only if $f(x_n) \rightarrow f(y)$ implies $x_n \rightarrow y$.

Whereas bijectivity just tests whether distinct points are mapped to distinct points, being a homeomorphism means nearby points must be mapped to nearby points, in both directions, from $X$ to $Y$ and from $Y$ to $X$. (That is, a homeomorphism preserves the topology.)

This suggests the following example. Let $X$ be the interval $[0,2\pi) \subset \mathbb{R}$ and let $Y$ be the unit circle $Y = \{ (x_1,x_2) \in \mathbb{R}^2 \mid x_1^2 + x_2^2 = 1\}$. Let $f\colon X \rightarrow Y$ be the function sending $\theta \in [0,2\pi)$ to $(\cos\theta,\sin\theta)$. Then $f$ is continuous because it sends nearby points in $X$ to nearby points in $Y$. But $f^{-1}$ is not continuous because the point $(1,0) \in Y$ is sent to the point $0 \in X$ but a point just below $(1,0) \in Y$, say $(\cos\theta,\sin\theta)$ for $\theta$ just less than $2\pi$, is sent to a point far away from $0 \in X$. (The reader should pause to understand this fully; drawing a picture may help.)

The above example is quintessential; an injective map $f$ is a homeomorphism only if it is not possible to map a “line $\longrightarrow$” to a “loop $\circlearrowleft$“. In detail, let $x_n$ be a sequence of points that either does not have a limit point, or that converges to a point $x$ that is different from $x_1$. Such a sequence can be obtained by considering the location at regularly spaced intervals of an ant walking along a path in $X$ that does not eventually return to where it started. It is therefore referred to loosely as a “line”. Let it loosely be said that this “line” gets mapped to a “loop” if $\lim f(x_n) = f(x_1)$. If it is possible for a line to be mapped to a loop, then $f$ cannot be a homeomorphism. (Draw some pictures!)

To summarise, injectivity of $f$ means that if $f(x) = f(y)$ then  $x = y$. For $f$ to be a homeomorphism requires more: if $\lim f(x_n) = f(y)$ then it must be that $\lim x_n = y$. Whereas injectivity just means the preimages of distinct points are distinct, a homeomorphism means the preimages of  convergent sequences with distinct limits are convergent sequences with distinct limits. Perhaps the best way of visualising this though is asking whether a “line” can be mapped to a “loop”.

Relevance of the Domain being Compact and the Codomain being Hausdorff

Let $x_n$ be a “line”, as defined above. There are two cases to consider: either $x_n$ does not have a limit point, in which case it will be seen that compactness of $X$ is relevant, or $x_n$ has a limit, in which case it will be seen that $Y$ being Hausdorff is significant.

When dealing with sequences, the role of compactness is guaranteeing every sequence has a limit point (or equivalently, that a convergent subsequence exists). Therefore, if $X$ is compact, it is not possible for $x_n$ to lack a limit point, and this option is immediately off the table.

If it is possible for a “line” to be mapped to a “loop” despite $X$ being compact then it is possible to achieve this assuming $x_n$ has a limit (by considering a convergent subsequence if necessary). Henceforth, assume $x_n \rightarrow x \neq x_1$ yet $f(x_n) \rightarrow f(x_1)$. Continuity of $f$ implies $f(x_n) \rightarrow f(x)$. When dealing with sequences, the role of the Hausdorff requirement is to ensure limits are unique. (Prove this for yourself, or look up a proof that limits are unique in Hausdorff spaces.) Therefore, $Y$ being Hausdorff means $f(x) = f(x_1)$ since $f(x_n)$ converges to both $f(x)$ and $f(x_1)$. Since $f$ is injective, this means $x = x_1$, contradicting $x_n$ being a “line”. Therefore, it is impossible for a “line” to be mapped to a “loop”; every injective map $f$ from a compact space to a Hausdorff space is automatically a homeomorphism.

The above has hopefully taken away a bit of the mystery of why compactness and the Hausdorff condition enter into the picture.

• An injective map $f$ (on a nice topological space where it suffices to reason with sequences) is a homeomorphism if and only if it is not possible for a “line $\longrightarrow$” to be mapped to a “loop $\circlearrowleft$“. (A line and a loop are topologically distinct whereas the purpose of a homeomorphism is to preserve topological structure.)
• If $y_n$ is a loop in $Y$, that is, $y_n \rightarrow y_1$, then consider its preimage $f^{-1}(y_n)$. Unless there is some reason for $f^{-1}(y_n)$ to converge, the fact that $f$ is injective is powerless to prevent $f^{-1}(y_n)$ from being a “line”, i.e., it may well be the case that $f$ is not a homeomorphism.
• If $X$ is compact then $f^{-1}(y_n)$ has a limit point. By considering a subsequence if necessary, assume without loss of generality that $f^{-1}(y_n) \rightarrow x$. Continuity of $f$ implies $y_n \rightarrow f(x)$.
• If $Y$ is Hausdorff then limits are unique, therefore, $f(x) = y_1$. Therefore, the preimage of a loop is a loop, not a line. The topology has been preserved, hence $f$ is a homeomorphism.
• The above is “intuitively” why the result holds, yet the formal proof based on open sets is shorter, more general but devoid of intuition (at least for beginners); this is the joy of mathematics.

## Morphisms

Very briefly, it is remarked that structure-preserving functions are called morphisms (in category theory). Two sets are equivalent if there exists a bijective function between them. As more structure is given to these sets, requirements stronger than just bijectivity are necessary if the structure is to be preserved.

In linear algebra, two vector spaces are the same if there exists a bijective function $f$ between them that is linear and whose inverse is linear; the linearity of $f$ and $f^{-1}$ is what preserves the structure. Interestingly, if $f$ is linear and bijective then its inverse $f^{-1}$ is automatically linear.

In topology, a continuous function preserves structure in one direction, therefore, to preserve structure in both directions requires a bijective $f$ that is continuous and whose inverse $f^{-1}$ is also continuous; as introduced earlier, such a function is called a homeomorphism. Unlike the linear case, $f$ being bijective and continuous is not enough to imply $f^{-1}$ is continuous.

1. July 25, 2014 at 2:03 am

Had missed this in my earlier perusal of your blog, nice exposition. A lingering doubt, probably silly, but could that compactness requirement be relaxed to closedness. Although, for good topological spaces, compactness is equivalent to completeness and totally boundedness. Can one ‘intuit’ this result in terms of these concepts separately?

• July 25, 2014 at 8:53 am

Excellent questions. The space X being closed is not enough; the real line is closed yet the real line can be mapped to a circle so that the limit is the same whether x goes to positive infinity or negative infinity. (Equivalently, the real line is homeomorphic to the open interval $(-\pi,\pi)$ and this open interval maps to the circle by $\theta \mapsto (\cos\theta,\sin\theta)$.)

The concepts of completeness and total boundedness require more structure than just a topology before they can be defined. The simplest is to assume the space is a metric space. In this case, it is intuitively clear why every sequence has a convergent subsequence: given a sequence, break the space up into a finite number of small regions. At least one region must contain an infinite number of terms of the sequence. Limit attention just to this region. Break this region up into a finite number of even smaller regions and keep just a single region that contains an infinite number of terms of the sequence, and repeat, with the size of the regions shrinking to zero. Here, total boundedness is being used to ensure this is possible. Intuitively, the selected region shrinks down to just a point and this is the desired limit point to which a subsequence converges, provided of course no one has “removed” that limit point from the space. This is where completeness comes in: sequences that “should” have a limit, do have a limit.

• July 25, 2014 at 12:23 pm

aah…ok…yeah that makes perfect sense. I see both are necessary. By more structure you mean in general, something like a uniform space? Thanks a lot for clarifying, cheers.

2. July 25, 2014 at 12:41 pm

Yes, a uniform space would work. In terms of intuition though, sticking with metric spaces suffices. (Of course, a metric induces a topology: the topology is generated by open balls, and open balls are defined with respect to the metric.)

3. November 24, 2015 at 6:37 pm

Thank you for a great article! I was having trouble figuring this out for my topology class. The proof seemed reasonably simple to write, but the reasoning behind it was not as clear. This really helped me to understand.