Are symplectic maps immersions?

Question

Let $(M_1,\omega_1)$ and $(M_2,\omega_2)$ be sympletic manifolds. We define a sympletic map $\varphi :M_1 \rightarrow M_2$ such that for the pull back we have to following $\varphi^*(\omega_2)=\omega_1$.

I want to show that this implies $\varphi$ is an immersion. From $\varphi^*(\omega_2)=\omega_1$ we get that at $p\in M_2$ and $X,Y\in T_pM_2$, $(\varphi^*\omega_2)_p(X,Y)=(\omega_2)_p(d\varphi X,d\varphi Y)=\omega_2(X,Y)$.

To show $\varphi$ is an immersion we want to show that $d\varphi_p$ is injective for all $p$. It's clear we will have to invoke the non degeneracy of the symplectic forms but I am unsure how to go from $d\varphi$ to $d\varphi_p$.

Answer 1

You have got your definitions wrong. For $p \in M_1$, $X,Y \in T_pM_1$, $(\omega_1)_p(X,Y)=(\varphi^*\omega_2)_p(X,Y)=(\omega_2)_{\varphi(p)}(d_p\varphi(X),d_p\varphi(Y))$.

So if $d_p\varphi(X)=0$, then for all $Y \in T_pM_1$, then $(\omega_1)_p(X,Y)=0$ so $X=0$.