Let $M$ and $N$ be two finite dimensional and real smooth manifolds and $f:M\to N$ a local diffeomorphism at some point $p\in M$. Is the induced map $f_{\star p}:T_pM\to T_{f(p)}N$ an isomorphism?
if $f:M\to N$ is local diffeomorphism implies $f_{\star p}:T_pM\to T_{f(p)}N$ is an isomorphism
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smooth-manifolds
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0Do you know the inverse function theorem? – 2017-02-07
2 Answers
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Diffeomorphism means $ \exists g : N \to M$ such that g is also differentiable and inverse of f. Thus $ (g \circ f ) _* = g_* \circ f_* = 1_{T_pM} $ Also $ ( f \circ g)_* = f_* \circ g_* = 1_{T_{f(p)}N}$ So $ f_*$ and $g_*$ are inverses on the vector spaces and hence an isomorphism.
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0f is a **local** diffeomorphism. – 2017-02-07
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0The map f is not a diffeomorphism, but a local diffeomorphism. – 2017-02-07
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1Then $f$ has a local differentiable inverse, and the same argument applies. – 2017-02-07
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Are you aware of the following facts?
$1)$ If $p$ is contained in an open subset $U\subseteq M$, then the inclusion map $i:U\to M$ induces an isomorphism $T_p U\cong T_p M$.
$2)$ If $F:M'\to N'$ is a diffeomorphism, then $(F_{*})_p:T_p M'\to T_p N'$ is an isomorphism.
If so, then you just need to put these together along with the fact that you get neighborhoods $U$ of $p$ and $V$ of $f(p)$ such that $f|_U:U\to V$ is a diffeomorphism.
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1The OP is asking essentially for the proof of (2). – 2017-02-07