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Let $f\colon M \rightarrow N$ be a smooth map. If $f$ is a diffeomorphism I am trying to show that the linear map $f_*$ : $T_pM \rightarrow T_{f(p)}M$ is an isomorphism for all $p \in M$. I know the derivative map $T_pf\colon T_pM \rightarrow T_{f(p)}N$ is an isomorphism, but I have trouble to proving when the map defines $T_pM \rightarrow T_{f(p)}M$.

Thank you for your help.

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What's the difference between $T_pf$ and $f_\ast$? –  Sam Oct 11 '12 at 20:01
    
$f_*$ is a map to $T_{f(p)}M$ and $T_pf$ is a map to $T_{f(p)}N$ –  L.S. Oct 11 '12 at 20:05
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@Clement $T_{f(p)}M$ doesn't make sense, as $f(p) \in N$. Usually, $f_* = T_pf$ are just different notiations for the same map. –  martini Oct 11 '12 at 20:18
    
Another redundant answer, just to drive home what every else has said already. The mapping you talk about, called the push-forward of $ f $, takes a tangent vector at $ p $ to $ M $ and associates it with that at $ f(p) $ to $ N $. So the $ T_p(f) $ you speak about doesn't make sense as that point $ f(p) $ cannot be in $ M $ unless $ f: M \rightarrow M $. If that is the case then probably your question would make sense. –  Vishesh Oct 12 '12 at 11:24

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