# Understanding the equivariant rank theorem

The statement: Let $M$ and $N$ be smooth manifolds, and let $G$ be a Lie group. Suppose $F:M\rightarrow N$ is a smooth map that is equivariant with respect to a transitive smooth $G$-action on $M$, and any smooth $G$-action on $N$. Then $F$ has constant rank.

Proof: Let $\theta$ and $\phi$ denote the $G$-actions on $M$ and $N$, respectively, and choose $p_0\in M$. For an arbitrary point $p$ we can find a $g$ so that $\theta_g(p_0)=p$. By equivariance, we have $\phi_g\circ F=F\circ\phi_g$, so applying the pushforward functor, we get the following commutative diagram:


The proof then says that $\theta_{g*}$ and $\phi_{g*}$ are linear isomorphisms, which is where I'm confused. Does this follow from the fact that $\theta_g$ and $\phi_g$ are diffeomorphisms? If so, is it true in general, i.e. if $\varphi:M\rightarrow N$ is a diffeomorphism, is $\varphi_*:T_pM\rightarrow T_{\varphi(p)}N$ an isomorphism?

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Wait, maybe this is a silly question. In the general case we know $T_pM$ and $T_{\varphi(p)}N$ have the same dimension, so we can just map the basis vectors for one bijectively to the basis vectors of the other? –  Sid Raval Feb 20 '12 at 3:40
This is true in general just because the pushforward is a covariant functor. If $\phi$ is a diffeomorphism, then let $\psi$ be its inverse. Then $\phi\circ \psi = id$, but now apply pushforward to get $\phi_*\circ \psi_* = id_*=id$, and likewise for the reverse inclusion to get that $\phi_*$ is an isomorphism. –  Matt Feb 20 '12 at 5:22
I should also point out that even though what you say is true, you can map $T_pM$ isomorphically to $T_{\phi (p)}N$ just by a bijection of basis vectors (diffeomorphisms preserve dimension) it only says that they are "abstractly isomorphic" and it doesn't tell you that the map $\phi_*$ is an isomorphism. –  Matt Feb 20 '12 at 5:25