Suppose I have two homeomorphisms $f:X\to Y$ and $g:X'\to Y'$. I'm trying to show that $h:X\times X'\to Y\times Y'$ defined by $h(x,x')=(f(x),g(x'))$, is a homeomorphism. I think I've got it mostly worked out, but I have a couple little uncertainties.
I've established that $h$ is injective, and I'm trying to prove that $h$ is a homeomorphism by showing that $h^{-1}(B)$ is open for every basis element $B$ of $Y\times Y'$.
@tb Right, I'd have to prove the same thing for $h^{-1}$ to establish that $h$ and $h^{-1}$ are both continuous, therefore implying that $h$ is a homeomorphism.
@DavidK: It is enough to show that $f \times g$ is continuous whenever $f$ and $g$ are, and to show that $(f \circ f') \times (g \circ g') = (f \times g) \circ (f' \times g')$ when these compositions make sense.
@tb Ok. If $B\in\mathcal{B}$ where $\mathcal{B}$ is a basis for a topology on $Y\times Y'$. I can then write $B=U\times U'$ where $U$ and $U'$ are open in $Y$ and $Y'$ respectively. Right?
@tb Yes. So I want to then prove that $h^{-1}(U\times U')$ is open in $X\times X'$ to establish that $h$ is continuous. I know that $f^{-1}(U)$ and $g^{-1}(U')$ are open in $X$ and $X'$ respectively. So $f^{-1}(U)\times g^{-1}(U')$ is open in $X\times X'$. I guess I'm unsure about $h^{-1}(U\times U')$.
Is $h^{-1}(U\times U')=f^{-1}(U)\times g^{-1}(U')$?
@ZhenLin I'm not clear as to how that helps establish a homeomorphism. But, we literally just covered the definition of a homeomorphism, and we have established no theorems at this point.
@ZhenLin Oh wait. I think I see how I can use that.
@ZhenLin I'm not sure I understand the thrust of Qiaochu's last paragraph here. Does he want to look as representations as enriched functors $G \to \operatorname{End}{(V)}$ instead of the "action" definition $G \times V \to V$?
I still don't understand what he is saying (or how he's not contradicting his previous paragraph). It's exactly the problem that Top is not enriched over itself that gives the wrong definitions when passing from the action setting to the functor setting.
