If the pullback of $F:M \to N$ is an isomorphism, is $F$ a diffeomorphism?

Question

Let $M,N$ be smooth manifolds, and let $F:M \to N$ be a smooth map. Then $F$ induces a map $F^*: C^\infty(N) \to C^\infty(M)$, given by $F^*(f)=f \circ F$. Suppose that $F^*$ is an isomorphism.

Q1: Is $F$ a bijection?

Q2: Is $F$ a homeomorphism? (This would imply that $F$ is a diffeomorphism.)

Answer 1

The answer to both your questions is "yes".

a) The surjectivity of $F^*$ already implies that $F:M\to N$ is a closed $C^\infty $ embedding.
This is not very difficult and a proof can be found on page 24 of that book.

b) If $F$ were not surjective, you could take any point $n\in N\setminus F(M)$ outside the image of $F$ and construct a bump function $f\in C^\infty (N)$ with $f(n)=1$ and with support disjoint from the closed submanifold $F(M)$.
Then the equalities $F^*(f)=F^*(0)=0\in C^\infty (N)$ would violate the injectivity of $F^*$.
This contradiction proves that $F$ is in reality surjective and, being also a closed embedding, is in fact a diffeomorphism.

Answer 2

Let's prove that $F$ is injective. Suppose by contradiction $F(x)=F(y)$. Then $F^*(g)(x) = g(F(x)) = g(F(y)) = F^*(g)(y)$. But there exists $f\in C^\infty(M)$ such that $f(x)\neq f(y)$ so $F^*(g)\neq f$ and this is true for every $g\in C^\infty(N)$. Hence $F^*$ is not surjective.