
Homeomorphisms
Let $\displaystyle f: (M_1,d_1)\to(M_2,d_2)$ and $\displaystyle G(f)=\{(x,f(x))x\in M_1\}\subseteq M_1\times M_2$ and let $\displaystyle \tilde f: (M_1,d_1)\to(G(f),d_p).$ Consider $\displaystyle \tilde f(x)=(x,f(x))$ and $\displaystyle d_p((x_1,x_2),(y_1,y_2))=\sqrt{d_1(x_1,y_1)^2+d_2( x_2,y_2)^2}$ for all $\displaystyle (x_1,y_1),(x_2,y_2)\in M_1\times M_2.$ Prove that $\displaystyle \tilde f$ is a homeomorphism.
In order to be a homeomorphism we need bijectivity for $\displaystyle \tilde f$ and continuity for $\displaystyle \tilde f$ and $\displaystyle \tilde f^{1},$ is there a faster way to solve this?

Re: Homeomorphisms
$\displaystyle \text{Let } \pi : (G(f), d_p) \rightarrow (M_1, d_1) \text{ by } \pi(x,y) = x \ \forall (x,y) \in G(f)$.
$\displaystyle \text{To show it's a bijection, show } (\tilde{f} \circ \pi)(x,y) = (x,y) \ \forall \ (x,y) \in G(f),$
$\displaystyle \text{and } (\pi \circ \tilde{f})(x) = x \ \forall \ x \in M_1.$
$\displaystyle \text{Note that } (G(f), d_p) \text{ is the subspace with the induced metric from the Euclideantype product metric } d_p^*$
$\displaystyle \text{on }(M_1, d_1) \times (M_2, d_2) \cong (M_1 \times M_2, d_p^*). \text{(Perhaps you'd want to prove the identity is a homeo?)}$
$\displaystyle \text{Thus } \pi = proj_1G(f) \text{ of } proj_1 : (M_1 \times M_2, d_p^*) \rightarrow (M_1, d_1), \text{ and so } \pi \text{ is continuous.}$
$\displaystyle \text{Note that } \tilde{f} \text{ is the subspace restriction of a composition of the diagonal map (always continous) with a continuous map: }$
$\displaystyle \Delta \ : \ (M_1, d_1) \rightarrow (M_1, d_1) \times (M_1, d_1), \ x \mapsto (x, x)$
$\displaystyle id \times f \ : \ (M_1, d_1) \times (M_1, d_1) \rightarrow (M_1, d_1) \times (M_2, d_2) \cong (M_1 \times M_2, d_p^*),$
$\displaystyle (id \times f)(x,y) = (x, f(y)).$
$\displaystyle \text{This means that } \tilde{f} = (id \times f) \circ \Delta \text{ is continuous.}$
$\displaystyle \text{If you're unconvinced, direct proofs for any part of the justification of continuity of } \pi \text{ and } \tilde{f} \text{ will be straightforward.}$