Suppose that the metric space $(X_i,d_i)$ is topologically equivalent to $(Y_i,d'_i)$ for $i=1,2, \cdots , n$. Show that the product metric spaces $X = \prod_{i=1}^nX_i$ and $Y= \prod_{i=1}^nY_i$ are topologically equivalent.

I know that since $(X_i,d_i)$ is topologically equivalent to $(Y_i,d'_i)$ for $i=1, \cdots , n$, then there exists a homeomorphism $f: (X_i,d_i) \to (Y_i,d'_i)$ from $(X_i, d_i)$ onto $(Y_i,d'_i)$ for which $f$ and its inverse $f^{-1}$ are both continuous. This is the definition of topological equivalence given in my book.

However, I am confused as to which homeomorphism to choose. I am given that it already exists, so do I just pick one, verify it is onto and one-to-one, and check for continuity of itself and its inverse? If this is the case, then how do I define this function in the first place?

Thank you very much for your time. Any suggestions and hints are most certainly appreciated. By the way, the textbook being used here is Principles of Topology by Fred H. Croom.

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    $\begingroup$ If $f_i:(X_i,d_i)\to (Y_i,d'_i)$ is the homeomorphism, $i=1,\dots,n$, then there is a canonical map $f:X\to Y$ induced by these $f_i$, and this map will be a homeomorphism. $\endgroup$ – Stefan Hamcke Feb 18 '15 at 15:45
  • $\begingroup$ In order to show that this canonical map is a homeomorphism, wouldn't I need some assignment in order to check for surjectivity and injectivity? $\endgroup$ – Jamil_V Feb 18 '15 at 16:02
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    $\begingroup$ Well, the map $f$ sends $(x_1,\dots, x_n)$ to the tuple $(f_1(x_1),\dots, f_n(x_n))$. Since you have inverse maps $g_i:Y_i\to X_i$, you can define $g:Y\to X$ in the same way, and then check that $gf$ is the identity on $X$. Of course you can also check surjectivity and injectivity, it is useful to know that surjectivity (resp. injectivity) of $f$ needs only surjectivity (resp. injectivity) of the functions $f_i$. But in your case checking that $g\circ f=id_X$ is possibly the shortest way to prove that $f$ is a homeomorphism. $\endgroup$ – Stefan Hamcke Feb 18 '15 at 16:11
  • $\begingroup$ What we make use of here is the universal property of the product space: Given topological spaces $(X_i)_i$ for indices in some (possibly infinite) set $I$, the product space with the product topology $X$ has continuous projections $p_i:X\to X_i$, and for any collection of maps $f_i:Y\to X_i$, there is a unique continuous map $f:Y\to X$ such that $p_i\circ f= f_i$. It seems like you are using some definition of "product metric space", so one obstacle to solving your problem might be in showing that the product metric induces in fact the product topology. $\endgroup$ – Stefan Hamcke Feb 18 '15 at 17:32

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