Projection is an open map

Let $X$ and $Y$ be (any) topological spaces. Show that the projection

$\pi_1$ : $X\times Y\to X$

is an open map.


Let $U\subseteq X\times Y$ be open. Then, by definition of the product topology, $U$ is a union of finite intersections of sets of the form $\pi_X^{-1}(V)=V\times Y$ and $\pi_Y^{-1}(W)=X\times W$ for $V\subseteq X$ and $W\subseteq Y$ open. This means (in this case) that we may without loss of generality assume $U=V\times W$. Now, clearly, $\pi_X(U)=V$ is open.

Edit I will explain why I assume $U=V\times W$. In general, we know that $U=\bigcup_{i\in I} \bigcap_{j\in J_i} V_{ij}\times W_{ij}$ with $I$ possibly infinite, each $J_i$ a finite set and $V_{ij}\subseteq X$ as well as $W_{ij}\subseteq Y$ open. Note that we have
(V_1\times W_1)\cap (V_2\times W_2) &= \{ (v,w) \mid v\in V_1, v\in V_2, w\in W_1, w\in W_2 \} \\&= (V_1\cap V_2)\times (W_1\cap W_2)
and this generalizes to arbitrary finite intersections. Now, we have
\pi_X(U)&=\pi_X\left(\bigcup_{i\in I}~ \bigcap_{j\in J_i} V_{ij}\times W_{ij}\right)
=\bigcup_{i\in I}~ \pi_X\left(\left(\bigcap_{j\in J_i} V_{ij}\right)\times \left(\bigcap_{j\in J_i} W_{ij}\right)\right)
= \bigcup_{i\in I}~ \bigcap_{j\in J_i} V_{ij} =: V
and $V\subseteq X$ is open, because it is a union of finite intersection of open sets. Note for the first equality also that forming the image under any map commutes with unions.

Source : Link , Question Author : Alisha , Answer Author : Jesko Hüttenhain

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