# 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.

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What have you tried? –  Davide Giraudo Nov 29 '12 at 20:27
How do you define the topology on $X\times Y$? –  Lior B-S Nov 29 '12 at 20:29
ProofWiki: Projection from Product Topology is Open –  Martin Sleziak Apr 11 '13 at 8:24
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. Now, we have $$\pi_X(U)=\pi_X\left(\bigcup_{i\in I}~ \bigcap_{j\in J_i} V_{ij}\times W_{ij}\right) =\bigcup_{i\in I}~ \bigcap_{j\in J_i} \pi_X( V_{ij}\times W_{ij}) = \bigcup_{i\in I}~ \bigcap_{j\in J_i} V_{ij} =: V$$ and $V\subseteq X$ is open, because it is a union of finite intersection of open sets. So basically, I made the assumption because forming the image under any map commutes with unions and intersections.
But I couldn't follow the very last step - would you please tell me why we can assume $U = V \times W$ given the previous construction? Thank you~ –  WishingFish Jul 5 '13 at 6:08
If $V_{ij}$ and $W_{ij}$ are arbitrary open, you don't need any intersections... (Intersection of rectangles is a rectangle.) And forming an image does not commute with intersections. –  tomasz Jul 5 '13 at 11:02