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Let be $(M, g)$ and $(N, h)$ Riemann Manifolds. Suppose that $f \colon M \longrightarrow N$ is a smooth function such that $f^{*}h = g$. Prove that $f$ is a immersion.

I attempt to use the Riemannian metric and manifold properties such as locally finite compactness, but I have many doubts; there may be some trick to solve this question??

Sumanta
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Santiago
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    See here – Sumanta Jul 17 '21 at 16:13
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    Hi, welcome. Yes, you should use the metric. No, I don’t think locally finite compactness has anything to do with it. Can’t you just test the definition of immersion? – Matthew Leingang Jul 17 '21 at 16:17
  • The point is $h_{f(m)}\big(df_m(v),df_m(v)\big)=(f^*h)m(v,v)=g_m(v,v)>0$ for any $v\neq 0$ as $g_m$ is a metric. So, for any $v\neq 0$ we have $h{f(m)}\big(df_m(v),df_m(v)\big)>0$; now $h_{f(m)}$ is a metric implies $df_m(v)\neq 0\implies df_m$ is injective for any $m\in M$. – Sumanta Jul 17 '21 at 16:19

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