Composition of local diffeomorphisms is a local diffeomorphism

diffeomorphismdifferential-geometrysmooth-functionssmooth-manifolds

Let $F: M\rightarrow N$ , $G:N\rightarrow P$ be local diffeomorphisms, where $M,N,P$ are smooth manifolds. I would like to show that $G\circ F: M\rightarrow P$ is a local diffeomorphism.

My attempt:

Let $x\in M$. Since $F:M\rightarrow N$ is a local diffeomorphism, there exists an open set $U$ of $x$ such that $F(U)$ is open in $N$ and $F|_U: U\rightarrow F(U)$ is a diffeomorphism. Similarly, since $F(x)\in N$, there exists a neighborhood $V$ of $F(x)$ such that $G(V)$ is open in $P$ and $G|_V: V \rightarrow G(V)$ is a diffeomorphism.

I thought of considering the set: $F|_U^{-1}(F(U)\cap V)$, since this set is open in $U$.However, I have not gotten far. May I have hints?

Please do not use immersions

Best Answer

I think you basically have it. Call $F|_U^{-1}(F(U)\cap V)=W$, then $(G\circ F)|_W$ is a diffeomorphism, since the composition of diffeomorphisms is a diffeomorphism, and because diffeomorphisms behave well under restriction.

There is another way to see this: $F$ is a local diffeomorphism at $p\in M$ if and only if $DF_p:T_pM\to T_pN$ is a nonsingular linear transformation. So, by your assumptions $DF_p$ and $DG_{F(p)}$ are both nonsingular, hence $$ D(G\circ F)_p=DG_{F(p)}\circ DF_p$$ is also.

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