# Homeomorphisms

• October 18th 2012, 04:43 PM
Megus
Homeomorphisms
Let $f: (M_1,d_1)\to(M_2,d_2)$ and $G(f)=\{(x,f(x))|x\in M_1\}\subseteq M_1\times M_2$ and let $\tilde f: (M_1,d_1)\to(G(f),d_p).$ Consider $\tilde f(x)=(x,f(x))$ and $d_p((x_1,x_2),(y_1,y_2))=\sqrt{d_1(x_1,y_1)^2+d_2( x_2,y_2)^2}$ for all $(x_1,y_1),(x_2,y_2)\in M_1\times M_2.$ Prove that $\tilde f$ is a homeomorphism.

In order to be a homeomorphism we need bijectivity for $\tilde f$ and continuity for $\tilde f$ and $\tilde f^{-1},$ is there a faster way to solve this?
• October 18th 2012, 09:28 PM
johnsomeone
Re: Homeomorphisms
$\text{Let } \pi : (G(f), d_p) \rightarrow (M_1, d_1) \text{ by } \pi(x,y) = x \ \forall (x,y) \in G(f)$.

$\text{To show it's a bijection, show } (\tilde{f} \circ \pi)(x,y) = (x,y) \ \forall \ (x,y) \in G(f),$

$\text{and } (\pi \circ \tilde{f})(x) = x \ \forall \ x \in M_1.$

$\text{Note that } (G(f), d_p) \text{ is the subspace with the induced metric from the Euclidean-type product metric } d_p^*$

$\text{on }(M_1, d_1) \times (M_2, d_2) \cong (M_1 \times M_2, d_p^*). \text{(Perhaps you'd want to prove the identity is a homeo?)}$

$\text{Thus } \pi = proj_1|G(f) \text{ of } proj_1 : (M_1 \times M_2, d_p^*) \rightarrow (M_1, d_1), \text{ and so } \pi \text{ is continuous.}$

$\text{Note that } \tilde{f} \text{ is the subspace restriction of a composition of the diagonal map (always continous) with a continuous map: }$

$\Delta \ : \ (M_1, d_1) \rightarrow (M_1, d_1) \times (M_1, d_1), \ x \mapsto (x, x)$

$id \times f \ : \ (M_1, d_1) \times (M_1, d_1) \rightarrow (M_1, d_1) \times (M_2, d_2) \cong (M_1 \times M_2, d_p^*),$

$(id \times f)(x,y) = (x, f(y)).$

$\text{This means that } \tilde{f} = (id \times f) \circ \Delta \text{ is continuous.}$

$\text{If you're unconvinced, direct proofs for any part of the justification of continuity of } \pi \text{ and } \tilde{f} \text{ will be straightforward.}$