2
$\begingroup$

Definition 1: A function $f:X\to Y$ for topological spaces $X$ and $Y$ is continuous if the preimage $f^{-1}(U)\subset X$ is open in $X$ for each $U\subset Y$ open in $Y$.

Definition 2: A function $f:X\to Y$ for topological spaces $X$ and $Y$ is open if a set $Z\subset X$ being open in $X$ implies that $f(Z)\subset Y$ is open in $Y$.

Definition 3: A function between topological spaces is a homeomorphism $\iff $ it is a continuous bijection and the inverse function $f^{-1}$ is also continuous.

Definition 4: Let $f:X\to Y$ be a function between sets. The preimage of a subset $U\subset Y$ is the collection of $x\in X$ such that $f(x)\in U$, notated $f^{-1}(U)$. Concretely, $$f^{-1}(U)=\{x\in X\mid f(x)\in U\}.$$

Theorem: The function $f:X\to Y$ for topological spaces $X$ and $Y$ is an open, continuous bijection $\iff$ $f$ is a homeomorphism.

Proof Attempt: Let $f:X\to Y$ be an open, continuous bijection. Our goal is to show that $f$ is a homeomorphism; in particular, we need to show that the inverse function $f^{-1}:Y\to X$ is continuous. To do this, we need to show that for each set $Q\subset X$ open in $X$ the preimage $(f^{-1})^{-1}(Q)\subset Y$ is open in $Y$. Let $Z\subset X$ be open in $X$. Consider the following manipulation of the definition of preimage: $$(f^{-1})^{-1}(Z)=\{y\in Y \mid f^{-1}(y)\in Z\}=\{y\in Y \mid y\in f(Z)\}=f(Z)\subset Y.$$ Since $f$ is open and $Z\subset X$ is open in $X$, $(f^{-1})^{-1}(Z)=f(Z)\subset Y$ is open in $Y$. Thus, $f^{-1}$ is continuous, and in particular $f$ is a homeomorphism.

Now let $f$ be a homeomorphism. Then $f$ is a continuous bijection with a continuous inverse $f^{-1}$. We aim to show that $f$ is open. Let $Z\subset X$ be open in $X$. Since $f^{-1}$ is continuous, the preimage $(f^{-1})^{-1}(Z)=f(Z)\subset Y$ is open in $Y$. Thus $f$ is open.$\;\;\;\boxed{}$

Notes: Does this proof make sense? Is it valid given the definitions? The equality $\{y\in Y \mid y\in f(Z)\}=f(Z)$ was very tricky for me to come to understand, and I'm only partly sure it is true.

Response to Answerers: Thank you for your time and expertise! I think I have a more economical grasp of the concept now. My newfound conception is that for a bijection $f$, $f$ being open is equivalent to $f^{-1}$ being continuous. This should be illustrated through the following diagram, where for notational ease I've relabeled $f^{-1}=g$. The key for me has been the understanding that the implications in the definitions of openness (of $f$) and continuity (of $g$) are the same: enter image description here

$\endgroup$
1
  • $\begingroup$ Well, glad we could help you understand. Nice drawing BTW. $\endgroup$ Oct 2 '21 at 17:29
1
$\begingroup$

You're overdoing the notation bit. The only observation of note is that for a bijection (!) being open is the same as the inverse being continuous.

To keep notation clean use $g: Y \to X$ for the unique inverse of $f$ (so that $f(x)=y$ iff $g(y)=x$ etc.) and note that indeed for any $O \subseteq X$ we have $$y \in g^{-1}[O] \iff g(y) \in O \iff \exists x \in O: g(y)=x \iff \exists x \in O: f(x)=y \iff y \in f[O]$$ hence

$$g^{-1}[O]=f[O]\tag{1}$$

from which it immediately follows that $f$ open implies $g$ continuous (taking $O$ to be any open set) or $f$ closed implies $g$ continuous (taking $O$ to be closed etc.) too, and vice versa.

$\endgroup$
1
$\begingroup$

Your proof is correct, but you've used more notation than is necessary, I think. The equality that you've expressed doubt about is a vacuous tautology - it's true, but in such a trivial manner that it's not really worth writing down (in particular, for any set $S \subseteq Y$, we have $S = \{y \in Y \mid y \in S\}$).

The logic of your proof is quite simple - ultimately, you're just observing that the "open map" condition is equivalent to continuity of the inverse ("the image of an open set is open" c.f. "the preimage of an open set is open"). So, maybe just say that, instead, and leave out the extra symbols?

$\endgroup$

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.