Let $X,Y,U,V$ be topological spaces, and $X$ is homeomorphic to $U$ and $Y$ is homeomorphic to $V$ Then is $X\times Y$ homemorphic to $U\times V$?

I have got the following maps

$n\colon X \rightarrow U$

$m\colon U \rightarrow X$

$s\colon Y \rightarrow V$

$t\colon v \rightarrow Y$

with $m\circ n \cong Id_X$

$n\circ m \cong Id_U$

$t\circ s \cong Id_Y$

$s\circ t \cong Id_V$

But I have no idea how to use these!

edit : so to show that it is a homeomorphism can I show that

$g \circ f(x,y) = (m \circ n(x), t \circ s(y))$ and can i then say that this is homeomorphic to $Id_{X \times Y}$ as $m \circ n$ is to $Id_X$ and $t \circ s$ is to $Id_Y$ ?

  • $\begingroup$ Hi - have you tried anything yourself? If you give any ideas that you have had, even if they haven't worked out, it would be useful, and you would likely get a lot more responses $\endgroup$ – Joe Tait Mar 12 '14 at 10:57
  • $\begingroup$ @JoeTait I have added an extra few lines, can i say that? $\endgroup$ – Math123456 Mar 12 '14 at 12:39
  • $\begingroup$ Please use LaTeX in your future questions: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Najib Idrissi Mar 12 '14 at 13:12

The map $f:X\times Y\to U\times V$ defined by

$f(x,y)=(n(x),s(y))$ is an homeomorphism with inverse $g(u,v)=(m(u),t(v))$.

  • $\begingroup$ i have added a few extra lines, is it sufficient to say that, and then for fog also? $\endgroup$ – Math123456 Mar 12 '14 at 12:55

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