30:06 I think that f has also to be shown to be continuous to conclude that f is an open map.

@anishagrawal332

2 ай бұрын

f is a bijective local homeomorphism, which means it is locally continuous, by a previous proposition, every locally continuous map is globally continuous, so we have that f is continuous.

Question: Continuity in topological sense, I understood that open sets goes to open sets, that is, if you take some open subset of the output set, then there is some open subset in the input set. So, is this 'open continouity' a way generalize injectivity?

@mariusfurter

2 жыл бұрын

That's an interesting thought. If you have a continuous function f: X -> Y, then any pre-image of an open set in Y is open in X by definition. So if you denote the set of opens in X by O(X) and similarly denote by O(Y) the opens in Y, then f induces a function O(Y) -> O(X) sending each subset of Y to its pre-image under f. On the other hand if f: X -> Y is an open map, then you get a function O(X) -> O(Y) sending each open set in X to its image under f. But this assignment does not have to be injective. For instance, the map f: R -> {0} sending each real number x in R to 0 is open with respect to the unique topology on the singleton set {0}, but clearly does not induce an injective map O(R) -> O({0}). If you were thinking about the induced function O(X) -> O(Y) being injective, this could be an interesting condition. I need to think about whether there are non-artificial cases where the induced function is injective, but where the original map is not.

@buraianmath

2 жыл бұрын

@@mariusfurter In the second paragraph did you mean O(Y)-> O(X)?

@mariusfurter

2 жыл бұрын

I was still thinking of the case where the map is open, but asking for the induced function O(Y) -> O(X) to be injective would work for any continuous map.

17:43 How do we know that f is continuous? You just showed that a closed bijective funtion f has a continuous inverse.

@mariusfurter

2 жыл бұрын

In the video I defined open / closed map to be *continuous* maps with certain properties. So the map f is assumed to be continuous when it is closed. This definition is non-standard and even the reference I'm using defines open / closed functions to be arbitrary functions which map open (closed) set to open (closed) sets. I must have mixed things up. Using the standard definition, we indeed would have to show that f is continuous when checking that it is a homeomorphism. In this case the equivalence only holds if we assume that f is a *continuous* bijection. The way the proposition is phrased could be interpreted in this way, since I say that f is a bijective map. When I say map I usually mean continuous function in the context of topological spaces, but I would have probably stressed this more if I actually had to use this fact. In any case, if you want to use the standard definition of open / closed function, the proposition still holds as long as you assume f is a continuous bijection.

@eamon_concannon

2 жыл бұрын

@@mariusfurter At 4:55 you do mention that f is a continuous bijective map. I missed that. Thanks a lot for the very helpful videos.

i guess that f need to be continuos to make a homeomorphism

@mariusfurter

2 жыл бұрын

Yes, homeomorphisms are those continuous maps that have continuous inverses. Perhaps I didn't say this explicitly enough at some point. In the video I defined open / closed maps to be *continuous* maps with special properties. I now see that this is non-standard, and people generally define an open / closed map to be *any* function with the corresponding property. This is probably why I neglect certain assumptions (because I have them in the definition). Apologies if this caused any confusion.