# 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.

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Hi Jesko, nice answer, thanks. It is very helpful for a question I just asked. –  WishingFish Jul 5 '13 at 6:05
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
@WishingFish: See my edit. Hope this helps! –  Jesko Hüttenhain Jul 5 '13 at 10:58
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
Also, $\pi_X(V_{ij} \times W_{ij})$ equals $V_{ij}$ (in general) only in the case that $W_{ij}$ is inhabited. Else it is empty. –  Ingo Blechschmidt May 22 '14 at 21:03

I was working through this same problem and would like to share my solution since there are some issues with the other answer (and it wasn't accepted). Please feel free to point out any flaws, of course.

Let $U$ be an open set in $X\times Y$. Then $U$ is a union of finite intersections of elements of $$\mathcal S = \left\{\pi_1^{-1}(A) : A\text{ open in } X\right\} \cup \left\{\pi_2^{-1}(B) : B \text{ open in } Y\right\},$$ that is, $$U = \bigcup_{\alpha\in I}\bigcap_{i\in J_i} S_{\alpha, i}$$ where each $J_i$ is finite and each $S_{\alpha, i}$ is in $\mathcal S$. We can write each $S_{\alpha,i}=\pi_1^{-1}(V_{\alpha,i})\cap\pi_2^{-1}(W_{\alpha,i})$, where each $V_{\alpha, i}$ is open in $X$ and each $W_{\alpha,i}$ is open in $Y$ (allowing for the possibility that $V_{\alpha,i}=X$ or $W_{\alpha,i}=Y$). As $$\pi_1^{-1}(V_{\alpha,i})=V_{\alpha,i}\times Y \text{ and } \pi_2^{-1}(W_{\alpha,i})=X\times W_{\alpha,i}$$ it follows that $$\pi_1^{-1}(V_{\alpha,i})\cap\pi_2^{-1}(W_{\alpha,i}) = (V_{\alpha,i}\times Y)\cap (X\times W_{\alpha,i}) = V_{\alpha,i}\times W_{\alpha,i}.$$ Letting $V_\alpha=\bigcap_{i\in J_i} V_{\alpha,i}$ and $W_\alpha = \bigcap_{i\in J_i}W_{\alpha,i}$, we have $$U = \bigcup_{\alpha\in I}\bigcap_{i\in J_i} V_{\alpha,i}\cap W_{\alpha,i} = \bigcup_{\alpha\in I}V_\alpha\times W_\alpha,$$ where each $V_\alpha$ is open in $X$ and each $W_\alpha$ is open in $Y$. It follows that $$\pi_1(U) = \pi_1\left(\bigcup_{\alpha\in I}V_\alpha\times W\alpha \right) = \bigcup_{\alpha\in I}\pi_1(V_\alpha\times W_\alpha) = \bigcup_{\alpha\in I'}V_\alpha$$ (where $I' = \{\alpha \in I : W_\alpha\ne\varnothing\}$) is open in $X$. We conclude that $\pi_1$ is an open map.

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Some similar approach is the following: Let $\pi_1 :X \times Y \to X$ be the projection and assume $U \subset X \times Y$ is open.

We must show that $\pi_1(U)$ is open. For this let $x_0 \in \pi(U)$. Then $x_0 = \pi(a_0,b_0)$ for some pair $(a_0,b_0) \in U$. Since $(a_0,b_0) \in U$ we can find two opens $a_0 \in R$ and $b_0 \in S$ with $R \times S \subset U$. That means $R \subset \pi_1(U)$ and we have $x_0 \in R$.

Now, $\pi_1(U)$ is a union of opens.

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