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My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

I briefly learned about covering spaces in an algebraic topology book but did not study in too much detail. I would like to try to restate Example 1.10 without complex variables and covering spaces.

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  • The context is that Example 1.10 makes me question Example 1.9, which I ask about here. I actually have more to ask about Example 1.10, and I would prefer that I would not have to go back to covering spaces and that I could think about $S^1$ in $\mathbb R^2$ instead of in $\mathbb C$.

  • Note: I assume Example 1.10 is in fact correct even though $F$ is not injective. (I ask about this here, here and here.)

Is the following restatement correct?

  1. Let $N$ and $M$ be the unit circle $S^1$ this time in $\mathbb R^2$. Let $F: N \to M$ be this time given by $F(x,y)=(x^2-y^2,2xy)$.

  2. I make no mention of "2-sheeted covering space map".

  3. My understanding is that

    • 3a. no particular Riemannian metric for $M$ in the original Example 1.10 is given and

    • 3b. the one $M$ could get from $\mathbb R^2$ is pointed out merely as a possibility.

  4. For any (see (3)), Riemannian metric $\langle, \rangle$ on $M$, we have that for any $p \in N$ and $v,w \in T_pN$, $\langle v, w\rangle'_p := \langle F_{*,p}v, F_{*,p}w\rangle_{F(p)}$ defines an inner product on $T_pN$.

  5. From (4), we get an assignment $\langle, \rangle': N \to \bigcup_{p \in N} \{\text{inner products on} \ T_pN\}$

  6. $\langle, \rangle'$ is a Riemannian metric on $N$ if $F$ is smooth and if $\langle, \rangle'$ is smooth (as defined in Definition 1.5).

  7. $F$ and $\langle, \rangle'$ are (somehow; this is what I intend to ask about in another question assuming this restatement is correct) smooth here in the restatement as in the original Example 1.10.

  8. Therefore, by (6) and (7), $\langle, \rangle'$ is a Riemannian metric on $N$.

  9. Again, $F$ is metric-preserving but not an isometry since $F$ is still not a diffeomorphism since $F$ is still not injective.

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    This looks fine. It should be pretty obvious that $F$ is smooth: just look at the formula you wrote for $F$ (and if you knew complex variables, you wouldn't even need to rewrite the formula, you could immediately conclude smoothness from the formula $F(z)=z^2$). If you compute the correct formula for $\langle,\rangle'$, you'll also be able to look at it and immediately conclude smoothness. – Lee Mosher Jun 21 '19 at 14:34
  • @LeeMosher I made a mistake. I get that $F$ is smooth. It's the $\langle , \rangle'$'s smoothness I intend to ask about (in another question). Thanks a lot for the verification. –  Jun 26 '19 at 08:01

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