How to prove that a given set is a torus

[glmuelle]

Hi

I'm trying to solve this exercise

"Prove that if C is a circular cylinder with S_1 and S_2 as its boundary circles and S_1 and S_2 are identified by mapping them both homeomorphically onto some third circle S, giving a map f: S_1 \cup S_2 \rightarrow S then (C - S_1 \cup S_2) \cup S with the identification topology (induced by the function that identifies all the points in S_1 and S_2 with corresponding points in S) is a torus."

Is it enough to show that the topology on (C - S_1 \cup S_2) \cup S is the subspace topology of R^3?

Thanks for your help

 

[quasar987]

No, that is not enough. A point has the subspace topology on R^3 but a point is not a torus, is it?

The most direct way to prove that two topological spaces are the same (i.e are homeomorphic) is of course to construct an explicit homeomorphism between them. In this case, it should not be difficult to do so. Call T the usual torus in R^3 and T' the wannabe torus. I suggest that you find a continuous bijective map T'-->T. Since T' is compact and T is Hausdorff, it will follow that your map is a homeomorphism.

Also note that to show that your map is continuous, it suffices to show that it is continuous as a map from T' into R^3, since T has the subspace topology induced by R^3.

 

[glmuelle]

Thank you very much, that answered my question!

 

[Jamma]

I don't understand you're question, but it's late and I'm probably just being completely mad, but can't you identify the two circles on a cylinder in a way to produce a Klein bottle instead of a torus? If you could show that it could also be embedded into R^3 then it would indeed be a torus because the Klein bottle cannot be embedded into R^3, only R^4; maybe this is what you meant at the end of your post?