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I defined torus as quotient space of $\mathbb{R}^2$, and $\pi$ is a quotient map. Then for each $ v\in\mathbb{R}^2$ I took a neighborhood $U(v)=B_{1/3}(v)$ and looked on $\pi|U:U\to \pi(U)=V$ and proved that they are bijections. Now for every such $V$ we regard $\pi^{-1}$ as coordinate and I need to compute change of coordinates and prove that in such way torus is equipped with the structure of a differentiable manifold. How should I do it?

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1 Answer 1

What you have to do is to take two different neighbourhoods as the ones you defined, $U_i$ and $U_j$, and check that the following map $$\pi_i^{-1} \circ \pi_j:V_i \cap V_j \longrightarrow V_j \cap V_i$$ is a smooth map. Observe that this map goes from an open set in $\mathbb{R^3}$ to an open set in $\mathbb{R^3}$ (it is actually the same), so proving its smoothness is a matter of smoothness within the euclidean space, so you may use its tools, namely the Implicit Function Theorem and such.

By the way, remember to check that the open sets you defined cover the whole torus; else way you don't get an atlas and then there is no differentiable structure to be proven.

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Maybe it should be $$\pi_i\circ \pi_j^{-1}:V_i \cap V_j \longrightarrow V_j \cap V_i$$ ? –  StudentMath Oct 18 '12 at 10:55
    
Hmmm... first you take out from an euclidean open set up to an open set in the manifold, so the inverse goes first. Then from the manifold you go back to the euclidean space. Let me clarify that I assume $(\pi_i^{-1} \circ \pi_j)(x) = \pi_j(\pi_i^{-1}(x))$. Does this make any sense to you? Observe that $\pi_i^{-1}(V_i \cap V_j) \subseteq U_i$. –  busman Oct 18 '12 at 11:03
    
oh yes, sorry, I thought other way around –  StudentMath Oct 18 '12 at 11:05
    
I myself am not always sure of the way it goes, and I have to stop and think about it for a few secs, but this time I did it on purpose to make sure I was not wrong :) –  busman Oct 18 '12 at 11:38

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