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I am reading the John Lee's Introduction to smooth manifolds, Proposition 22.12 and stuck at showing that every smooth 1-form $\sigma : M \to T^{*}M$ is an immersion :

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I am trying to understand the underlined statement. Why can we deduce that $\sigma$ is a immersion from the above coordinate representation? Why the rank of the matrix of partial derivatives of the coordinate representation is $n:=\operatorname{dim}M$ ?
I think that it seems possible to prove and stuck at making proof rigorously. I think that I am unfamilier to calculate the rank of the jacobian matrix. Can anyone helps ?

Plantation
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1 Answers1

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Since $\pi \circ \sigma = Id_{M}$, their differentials satisfy $$ \forall p\in M,\quad (\pi_*)_{\sigma(p)} \circ (\sigma_*)_p = Id_{T_pM}. $$ In particular, for any $p\in M$, $(\sigma_*)_p$ is injective, which is the definition of $\sigma$ being an immersion.

If you really want to use a matrix representation of $(\sigma_*)_p$, then differentiating the equality you have underlined gives a matrix representation of the form $$ \begin{pmatrix} 1 & 0 & \cdots & 0 & * & \cdots & *\\ 0 & \ddots & \ddots & \vdots & \vdots & & \vdots \\ \vdots & \ddots & \ddots & 0 & \vdots & & \vdots \\ 0 & \cdots & 0 & 1 & * & \cdots & * \end{pmatrix} $$ which thus has maximal rank.

Didier
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