Theorem: Suppose that $f:X \rightarrow Y$ is one-to-one, surjective and continuous. If $X$ is compact, Then $f^{-1}:Y \rightarrow X$ is also continuous.

The proof for this theorem is pretty easy. We can show that the inverse of every close set in $X$ is closed in $Y$ (Am I right?)

Now I want to show that being compact is necessary. My example is:

$$f(t) = (\cos(t),\sin(t)), \forall t\in [0,2\pi)$$

$f$ is one-to-one, surjective and continuous. But $f^{-1}$ is not continuous.

Now I'm looking for more examples. Or maybe a set of infinite examples.

  • 3
    $\begingroup$ What about $X=Y$ as sets, but $X$ has a finer (e.g., discrete) topology? $\endgroup$ – Hagen von Eitzen Jun 13 '15 at 13:54
  • 1
    $\begingroup$ @HagenvonEitzen Yes,Yes. Thats a very good example. $\endgroup$ – lino Jun 13 '15 at 13:59
  • $\begingroup$ Isn't that statement false without $Y$ Hausdorff? E.g. pick $X = Y = \{0,1\}$, with $X$ discrete and $Y$ indiscrete, and $f=1_X$. $\endgroup$ – Alex Provost Jul 19 '16 at 20:38
  • $\begingroup$ @AlexProvost I think you are right. $\endgroup$ – 6005 Jul 19 '16 at 20:41

First, the theorem you state actually requires $Y$ to be Hausdorff as well. See Proofwiki for the proof in that case, and Alex Provost's comment for a counterexample if $Y$ is not Hausdorff.

What about examples where compactness is neessary? Examples of this exactly correspond to examples of topologies $(X, \tau)$ and $(X, \tau')$ such that $\tau \subsetneq \tau'$. Why? Since $f$ is a bijection between $X$ and $Y$, we may simply identify the points of $X$ and the points of $Y$. Then, the fact that $f$ is continuous says that every open set in $Y$ is an open set in $X$; to make its inverse not continuous, we simply require $X$ to have open sets that are not in $Y$. So $X$ has some topology, $\tau'$, and it must strictly contain $Y$'s topology $\tau$ ("contain" meaning under the identification given by the bijection).

If we also want the counterexample to have $\tau, \tau'$ Hausdorff, we should just ensure that $\tau$ is Hausdorff because any finer topology of a Hausdorff topology is Hausdorff.


  • Hagen von Eitzen's example is pretty good to start: take $\tau$ to be any topology you like, say the standard topology on $[0,1]$, and take $\tau'$ to be the discrete topology on that set.

  • Your example with $\sin(t), \cos(t)$ is, in disguise, the following example: Take $X = [0,1)$. Let $\tau'$ be the subspace topology on $X$ inhereted from $[0,1]$. On the other hand, let $\tau$ be the topology on $X$ from the quotient space $[0,1] / \{0,1\}$. Then $\tau \subsetneq \tau'$ as we wanted.

  • The weak topology $\tau$ on a (say, normed space) $X$ is strictly contained in the strong topology $\tau'$.

I'm sure there are many more examples.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.