Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: deprecated: topos theory

Topic: fiberwise homeomorphisms

Jens Hemelaer (Jan 14 2021 at 16:38):

If $f : E \to E'$ is a continuous map of étale spaces over some base topological space $B$, then it is an homeomorphism if and only if it is a fiberwise homeomorphism. (This is a translation of "a morphism of sheaves is an isomorphism if it is an isomorphism in each stalk".)

Are there any generalizations to this where the spaces $E$ and $E'$ are no longer étale? For example, is it enough to assume that the structure maps $\pi : E \to B$ and $\pi : E' \to B$ are both open?

In any case, if $f : E \to E'$ is a fiberwise homeomorphism, then it is a bijection. So the difficulty is in showing that $f$ is open.

Jens Hemelaer (Jan 15 2021 at 10:54):

If $E$ and $E'$ are both topological manifolds of the same dimension (not necessarily connected), then any continuous bijection $f : E \to E'$ is automatically a homeomorphism (by invariance of domains). So in this case, it is enough to check bijectivity, or equivalently, bijectivity in each fiber.

I was hoping for results that work for general topological spaces, but this already excludes some counterexamples.

Morgan Rogers (he/him) (Jan 16 2021 at 16:10):

If you already have some spatial counterexamples, that answers your question about open maps, since all spatial locales are open over the terminal locale :+1:

Jens Hemelaer (Jan 17 2021 at 10:33):

@Morgan Rogers (he/him) With "space" I meant "topological space", I'll edit what I wrote above. In any case, the implication "fiberwise homeomorphism implies homeomorphism" trivially holds over the singleton topological space.

Jens Hemelaer (Jan 17 2021 at 11:34):

I think I now have a counterexample thanks to the answers to a related question here:
https://math.stackexchange.com/q/3986209/

Let $Q = \{ (x,y) \in \mathbb{R}^2 : x \geq 0,~ y\geq 0\} - \{(0,0)\}$ be the upper right quadrant with the origin removed.
Consider the action $\mathbb{R} \times Q \to Q$ given by $\lambda \cdot (x,y) = (2^\lambda x, 2^{-\lambda} y)$.
Fold $Q$ into a Möbius strip $M$ by identifying $(t,0)$ with $(0,1/t)$.
The action of $\mathbb{R}$ on $Q$ descends to an $\mathbb{R}$-action on the Möbius strip $M$.
Now the quotient space is $M/\mathbb{R} \cong [0,\infty)$.
The quotient map $\rho : M \to [0,\infty)$ is open, and has a section $s$ given by $s : [0,\infty) \to M,~ s(t) = (1,t)$.

Now consider the map $f : \mathbb{R} \times [0,+\infty) \to M$ with $f(\lambda,t) = \lambda \cdot s(t)$.
This map satisfies $\rho \circ f = \pi_2$ and both $\rho$ and $\pi_2$ are open maps.
Restricted to a single fiber, $f$ is a bijection between two spaces that are both homeomorphic to $\mathbb{R}$, so by invariance of domains, $f$ is in each fiber an homeomorphism. However, $f$ itself can't be an homeomorphism, since the domain is contractible and the codomain isn't.

Together with @Joshua Wrigley I'm trying to find a nice characterization of "$G$-torsors" for $G$ a topological group, and the above gives an example of what can go wrong.

Morgan Rogers (he/him) (Jan 19 2021 at 12:01):

Jens Hemelaer said:

Now the quotient space is $M/\mathbb{R} \cong [0,\infty)$.

Can you explain how you arrived at this conclusion? I can't see where $0$ comes from, for example.

Morgan Rogers (he/him) (Jan 19 2021 at 12:06):

Ah, I suppose the $0$ comes from the identified axes, but since the $\mathbb{R}$ action can move any point arbitrarily close to one of these, shouldn't that end up being a dense point in the quotient?

Jens Hemelaer (Jan 19 2021 at 13:17):

Morgan Rogers (he/him) said:

Ah, I suppose the $0$ comes from the identified axes, but since the $\mathbb{R}$ action can move any point arbitrarily close to one of these, shouldn't that end up being a dense point in the quotient?

Oh right... I should double check this.

Jens Hemelaer (Jan 19 2021 at 13:24):

Let $f: M \to M/\mathbb{R}$ be the quotient map. For quotients by group actions, the quotient map is always open. So the open subsets of $M/\mathbb{R}$ are precisely the sets of the form $f(U)$ for $U \subseteq M$ open. Using this characterization, you can check that the topology on $M/\mathbb{R}$ agrees with the topology on $[0,+\infty)$.

Morgan Rogers (he/him) (Jan 19 2021 at 15:02):

I think that the fact we ignore the topology on $\mathbb{R}$ when considering the action but not in the product was confusing me. :grinning_face_with_smiling_eyes:

Morgan Rogers (he/him) (Jan 19 2021 at 15:05):

Jens Hemelaer said:

Now consider the map $f : \mathbb{R} \times [0,+\infty) \to M$ with $f(\lambda,t) = \lambda \cdot s(t)$.
This map satisfies $\rho \circ f = \pi_1$ and both $\rho$ and $\pi_1$ are open maps.

I still think something needs tweaking here, since when $\lambda = -1$, for example, you go outside the domain of definition of $M$. Maybe take $(0,+\infty)$ instead of $\mathbb{R}$ here?

Jens Hemelaer (Jan 19 2021 at 15:57):

The dot denotes the group action of $\mathbb{R}$ on $M$, not multiplication. Not the best notation on my part, and I don't see a good alternative...

Morgan Rogers (he/him) (Jan 19 2021 at 16:01):

Okay, that's all of my confusions resolved now, I think, as long as by $\pi_1$ you mean $2^{\pi_1}$? :rolling_on_the_floor_laughing:

Morgan Rogers (he/him) (Jan 19 2021 at 16:16):

What I like the most about this example is that if you try something involving a more typical Möbius strip, such as the 1-dimensional vector bundle over the circle, you can't construct the maps you would need to construct a counterexample. The map $[0,2\pi) \to S^1$ isn't open, for example, so the mapping $\mathbb{R} \times [0,2\pi) \to M$ can't be used as a counterexample.

Jens Hemelaer (Jan 19 2021 at 17:52):

Morgan Rogers (he/him) said:

Okay, that's all of my confusions resolved now, I think, as long as by $\pi_1$ you mean $2^{\pi_1}$? :rolling_on_the_floor_laughing:

Hmm, $\pi_1$ should have been $\pi_2$ :grimacing: thanks, I edited it!

Ming Ng (Jan 19 2021 at 20:16):

Jens Hemelaer said:

Together with Joshua Wrigley I'm trying to find a nice characterization of "$G$-torsors" for $G$ a topological group, and the above gives an example of what can go wrong.

Interesting. Just to make sure I got things right: could you be a little bit more explicit about the connection between $G$-torsors and this example of a fibrewise homeomorphism that is not a homeomorphism? I understand that you've defined a map $f$ between two spaces $\mathbb{R}\times [0,\infty)$ and $M$ over $[0,\infty)$, and I broadly understand (save for the details about $M/\mathbb{R}\cong [0,\infty)$) how your previous remarks show that $f$ is fibrewise homeomorphic. But what does this have to do with $G$-torsors exactly?

Is it that your example can also be interpreted as defining an $\mathbb{R}$-space, when we view it as $\rho\circ f$ defining an action $\mathbb{R}\times [0,\infty)\rightarrow [0,\infty)$? And that the failure of $f$ to be a homeomorphism relates to the failure of the desired induced map between $\mathbb{R}\times[0,\infty)\xrightarrow{\sim} [0,\infty)\times [0,\infty)$ to be an isomorphism (which we require for an $\mathbb{R}$-torsor)? But is it obvious that $M\cong [0,\infty)\times [0,\infty)$?

Jens Hemelaer (Jan 20 2021 at 09:21):

@Ming Ng If $E$ is a space with a $G$-action, together with a map $p : E \to B$ such that $p(g \cdot e) = p(e)$, then $E$ is a $G$-torsor if and only if the map
$\xi : G \times E \to E \times_B E,~ (g,x) \mapsto (gx,x)$
is a homeomorphism. It follows in particular that the action is free, and that the action is transitive on the fibers.
Further, there are natural projection maps $\pi_2 : G \times E \to E$ and $\pi_2': E \times_B E \to E$, and for these projection maps we have that $\pi_2' \circ \xi = \pi_2$. So $\xi$ is a map "over" $E$.

The question is now, is it enough that $\xi$ is a fiberwise homeomorphism to deduce that it must be a homeomorphism? This would simplify a lot, because on the fibers over some point $x \in E$, the map $\xi$ is given by the map
$\xi : G \to \mathcal{O}(x),~ g \mapsto g \cdot x$,
where $\mathcal{O}(x)$ denotes the orbit of $x$.

So in order for $\xi$ to be a fiberwise homeomorphism, it is enough that in $E$, each orbit is (in a natural way) homeomorphic to $G$!

Unfortunately, if a map is a fiberwise homeomorphism that doesn't necessarily mean it is a homeomorphism, see for example the map $f$ from above.

On the other hand, if $E$ and $E'$ are both $G$-torsors over $B$, and $g : E \to E'$ is any $G$-equivariant map over $B$, then I think $g$ is necessarily a homeomorphism (non-isomorphic torsors don't have maps between them). In the example above, $f : \mathbb{R} \times [0,+\infty) \to M$ is an $\mathbb{R}$-equivariant map over $[0,+\infty)$, which is not a homeomorphism. The domain is a (trivial) torsor over $[0,+\infty)$, so this means that the codomain can't be a torsor (despite the orbits being all naturally homeomorphic to $\mathbb{R}$).