I'm starting to study differential geometry and my starting point is the book "Introduction to Topological Manifolds". It seems to be an appropriate choice but it does not present any solutions so, if you are willing, I ask you to review my solution to Exercise A.16.

Exercise: Prove that

"A map f:M_1\rightarrow M_2 between metric spaces is continuous if and only if the inverse image of every open set is open."


We start off with the definitions.

- Given any set M and a distance function d:M\times M\rightarrow\mathbb{R}_{\geq 0} satisfying the symmetry, positivity and the triangle inequality properties we say that the pair (M,d) defines a metric space.

- A function is continuous if and only if

\forall x\in M_1\ \forall\epsilon>0\ \exists\delta>0\ d_1(x,y)<\delta\implies d_2(f(x),f(y))<\epsilon

- A set U is open if and only if

\forall x\in U\ \exists r>0\ d(x,y)<r\implies y\in U

- Given a function f:M_1\rightarrow M_2, the inverse image of a set W\subset M_2 is given by f^{-1}(W):=\{x\in M_1: f(x)=y\text{ for some }y\in W\}

Let us first prove that if f is continuous and U_2\subset M_2 is open then U_1:=f^{-1}(U_2)\subset M_1 is also open, i.e.

\forall x\in U_1\ \exists\ r>0 d_1(x,y)<r\implies y\in U_1

For any x\in U_1, there exists \epsilon>0 such that d_2(f(x),f(y))<\epsilon\implies f(y)\in U_2 because we assumed U_2 to be open. From the continuity of f we conclude that there exists \delta>0 such that d_1(x,y)<\delta\implies d_2(f(x),f(y))<\epsilon. From the definition of U_1, we have that y\in U_1. Therefore, for r=\delta we have that d(x,y)<r\implies y\in U.

Now, we prove that if U_2\subset M_2 is open and U_1:=f^{-1}(U_2)\subset M_1 is also open then f is continuous. Suppose that it is not, then the following holds

\exists x\in M_1\ \exists\epsilon>0\ \forall\delta>0\ d_1(x,y)>\delta\land d_2(f(x),f(y))\geq\epsilon

Therefore, for each y such that d_1(x,y)<\delta we have that y\in U_1, because U_1 is an open set. From the definition of U_1 we conclude that there exists z=f(y)\in U_2. Take y=x, then z = f(x) and d_2(f(x),f(x))=0 because of the properties of the distance function, but this contradicts d_2(f(x),f(y))\geq\epsilon and thus f must be continuous.