# If $F$ is diffeomorphism the linear map $F_*$ is an isomorphism?

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 $T_pM \rightarrow T_{f(p)}M$.

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
@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