# Are symplectic maps immersions?

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 $$d\varphi_p$$.

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 $$X=0$$.