Let $X$ and $Y$ be topological spaces with topologies $\tau_X$ and $\tau_Y$. So, we say the neigborhoods of $X \times Y$ are of the form $N = O \times O'$ where $O \in \tau_X$ and $O' \in \tau_Y$. I want to show $p_X : X \times Y \to X$ is continuous. So, taking an open set $O$ in $X$ we show $p_X^{-1} (O)$ is open in $X \times Y$. Can we conclude that $p_X^{-1}(O) = O \times Y $. But $O \times Y$ is a neighborhood, and hence open, of $X \times Y$. In other words, My question is: How can we show $p_X^{-1} (O)= O \times Y $?
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Yes; it’s just a matter of checking the definitions.
- If $p=\langle x,y\rangle\in O\times Y$, then $p_X(p)=x\in O$, and therefore $p\in p_X^{-1}[O]$.
- Conversely, if $p=\langle x,y\rangle\in p_X^{-1}[O]$, then $x=p_X(p)\in O$, and $p\in O\times Y$.
Brian M. Scott
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$p_X^{-1}(O) = \{(x,y)\in X\times Y\mid p_X(x,y)\in O\} = \{(x,y)\in X\times Y\mid x\in O\} = O\times Y$
Daniel Robert-Nicoud
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