4 ms·

The idea is that the red point is being identified with the entire line, not stretched. My math is a little rusty, but I think you can make a diagram like http

by panic 10y ago

The idea is that the red point is being identified with the entire line, not stretched. My math is a little rusty, but I think you can make a diagram like http://i.imgur.com/PchSkyI.png http://i.imgur.com/PchSkyI.png and use the fact that Q1 and Q2 are quotient maps to define g and show it's a homeomorphism. The argument goes something like: there's a universal property on quotient spaces X with quotient map Q which says that if F(a) = F(b) for all a and b in X such that a and b are identified (a ~ b), there's a unique continuous G such that F = G ∘ Q. So let's take a and b in the top left space. We have a homeomorphism f such that a ~ b if and only if f(a) ~ f(b). Since quotients take equivalent elements to equal elements, Q2(f(a)) = Q2(f(b)). Therefore there's a unique continuous g1 such that Q2 ∘ f = g1 ∘ Q1. We can apply the the same reasoning to Q1 ∘ f^-1 to get a unique continuous g2 such that Q1 ∘ f^-1 = g2 ∘ Q2. Furthermore, g1 and g2 are inverses: g1 ∘ g2 ∘ Q2 = g1 ∘ Q1 ∘ f^-1 = Q2 ∘ f ∘ f^-1 = Q2, and Q2 is surjective. Therefore g = g1 is a homeomorphism.

johncolanduoni 10y ago
This argument is almost there. The theorem you mentioned about the quotient requires that you are taking a topological quotient on the left of your diagram, which prescribes a specific topology for the quotient space[1] (which you have not demonstrated is the topology you are using). Let Q1: X -> Y. This is the topology such that the open sets for Y are precisely those for which their preimage via Q1 is open in X. If you think this sounds a lot like continuity, you're right, but it adds one crucial detail: continuity only requires that whatever open sets we have for Y, their preimage is open (so for example you can make Y have the topology of a point, and Q1 would still be continuous). For quotient maps, we require that any set in Y with an open preimage be open, which fixes exactly one topology. This makes sense, because otherwise the quotient could pick from a laundry list of (non-isomorphic) topologies which keep the quotient map continuous, breaking the topological invariance of the quotient map. [1]: https://en.wikipedia.org/wiki/Quotient_space_(topology) https://en.wikipedia.org/wiki/Quotient_space_(topology)

by panic 10y ago

johncolanduoni 10y ago

This argument is almost there. The theorem you mentioned about the quotient requires that you are taking a topological quotient on the left of your diagram, which prescribes a specific topology for the quotient space[1] (which you have not demonstrated is the topology you are using). Let Q1: X -> Y. This is the topology such that the open sets for Y are precisely those for which their preimage via Q1 is open in X. If you think this sounds a lot like continuity, you're right, but it adds one crucial detail: continuity only requires that whatever open sets we have for Y, their preimage is open (so for example you can make Y have the topology of a point, and Q1 would still be continuous). For quotient maps, we require that any set in Y with an open preimage be open, which fixes exactly one topology. This makes sense, because otherwise the quotient could pick from a laundry list of (non-isomorphic) topologies which keep the quotient map continuous, breaking the topological invariance of the quotient map. [1]: https://en.wikipedia.org/wiki/Quotient_space_(topology) https://en.wikipedia.org/wiki/Quotient_space_(topology)