Homeomorphism and identity

Let $\displaystyle (M_1,d_1),\ldots,(M_n,d_n),$ and $\displaystyle M=M_1\times\cdots\times M_n.$ Prove the following:

$\displaystyle 1_M : (M,d_e)\to(M,d_m)$ and $\displaystyle 1_M :M,d_e)\to(M,d_s)$ are homeomorphisms, where $\displaystyle d_e$ is the euclidian distance, $\displaystyle d_s=|x_1-y_1|+\cdots+|x_n-y_n|,$ and $\displaystyle d_m$ distance of the maximum.

$\displaystyle 1_M$ is supposed to be the identity function, but I'm confused how to work with.

$\displaystyle f( N,d)\to (M,d_e)$ is continuous iff $\displaystyle f_i( N,d)\to(M_i,d_i)$ is continuous for $\displaystyle 1\le i\le n.$

The left implication is easy because if each $\displaystyle f_i$ is continuous then clearly $\displaystyle f$ is continuous, but don't know how to prove the right implication.

Re: Homeomorphism and identity

Topologically, the meaning of "the identity map on X is a homeomorphism" is that two seemingly different topologies are actually the same topology.

$\displaystyle \text{Suppose } \tau_1 \text{ and } \tau_2 \text{ are two (seemingly) different topologies for } X. \text{ (Such as arising from two different metrics.)}$

$\displaystyle 1_X : (X, \tau_1) \rightarrow (X, \tau_2) \text{ is a homeomorphism implies that it, and its inverse (itself again), are continuous.}$

$\displaystyle 1_X \text{ continuous means that the inverse image of any open set in } (X, \tau_2) \text{ is open in } (X, \tau_1).$

$\displaystyle \text{But the inverse image under that map of a set is just itself. So } 1_X \text{ continuous means that:}$

$\displaystyle \text{If } U \in \tau_2, \text{ then } 1_X^{-1}(U) \in \tau_1. \text{ But } 1_X^{-1}(U) = U, \text{ so } U \in \tau_1.$

$\displaystyle \text{Thus } 1_X \text{ continuous implies that: if }U \in \tau_2, \text{ then } U \in \tau_1.$

$\displaystyle \text{Thus } 1_X \text{ continuous implies that } \tau_2 \subset \tau_1.$

$\displaystyle \text{Likewise, } 1_X^{-1} \text{ continuous implies that } \tau_1 \subset \tau_2.$

$\displaystyle \text{So } 1_X \text{ a homeomorphism implies that } \tau_2 \subset \tau_1 \text{ and } \tau_1 \subset \tau_2.$

$\displaystyle \text{So } 1_X \text{ a homeomorphism is just another way of saying that } \tau_1 = \tau_2.$

In your case, you'll have different metrics, but you'll show that the identity is a homeomorphism, meaning that despite the different ways of measuring distances, the topologies those metrics induce (via their open balls) are the same. The way you show that is to show that every open ball in one metric is open in the topology induced by the other metric (but perhaps not an open BALL anymore), and visa versa.