# Continuity of a composite map between smooth manifolds

Let $M,N,P$ be smooth manifolds, $F\in\mathcal{C}^\infty(M,N)$ a smooth map and $i\in\mathcal{C}^\infty(P,N)$ an injective immersion, such that $F(M)\subset i(P)$. We define a map $G:M\rightarrow P$ by $F(p)=i(G(p))$.

I am to show the following:

If $i$ is an embedding, then $G$ is continuous.

My ideas so far:

Let $i$ be an embedding, then $i:P\rightarrow i(P)$ is a homeomorphism, i.e. the inverse $i^{-1}$ exists and is also continuous. Therefore, from $F=G\circ i$ follows that $G=F\circ i^{-1}$. This is well-defined since $F(M)\subset i(P)$.

Now one can pick an open subset $U\subset P$, such that $i(U)\subset F(M)$. Of course, $i(U)$ is open. But the question is how to show that $F^{-1}(i(U))=G^{-1}(U)$ is open in $M$.

Any hints or ideas? I don't see how injectivity or differentials would help me here, except that I know: If a smooth map between manifolds is a diffeomorphism, then its differential is a bijection. However, the converse is not always true.

• In my answer I overlooked the assumption $F(M)\subset i(P)$, so I deleted it. But the statement that you get the composition wrong is still correct, you have $F=i\circ G$, not $= G\circ i$, siimilarly $G= i^{-1}\circ F$. – Thomas Jun 5 '18 at 16:23
• Why can you pick an open set $U\subset P$ such that $i(U)\subset F(M)$? $F(M)$ could be a single point. – Thomas Jun 5 '18 at 16:25

Let $U$ be open in $P$, then $i(p)$ is open in $i(N)$. This means there is an open set $V\subset N$ such that $i(U)= V\cap i(N)$. Then $F^{-1}(V) = F^{-1}(i(U))$ is open since $F$ is continuous and $F(M) \subset i(N)$.