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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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4 Answers 4

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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 \begin{align*} (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) \end{align*} and this generalizes to arbitrary finite intersections. Now, we have \begin{align*} \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 \end{align*} 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.

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    $\begingroup$ 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. $\endgroup$
    – tomasz
    Jul 5, 2013 at 11:02
  • $\begingroup$ 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. $\endgroup$ May 22, 2014 at 21:03
  • $\begingroup$ That's wrong, $f(U\cap V)\subset f(U)\cap f(V)$ generally. See the very page you have linked. $\endgroup$
    – GFR
    Jul 5, 2015 at 19:39
  • $\begingroup$ @GFR: Indeed, that was a serious blunder. Should be fixed now. $\endgroup$ Jul 5, 2015 at 19:59
  • $\begingroup$ Think so, I have removed my down vote. You may want to edit your last line as well. $\endgroup$
    – GFR
    Jul 5, 2015 at 20:10
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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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The projection map

$\pi$ : $X\times Y\to X$ is defined by $\pi(A×B)=A$.

Suppose $U$ is open in $X×Y$. We need to prove that $\pi (U)$ is open in $X$. Since $U$ is open in $X×Y$ there exist basis element $A$ containing a point $x\in X$ and $B$ containing a point $y\in Y$ in $X$ and $Y$ respectively. Therefore $$x\in A=\pi(A×B)\subset\pi (U)$$

Thus $\pi U$ is open in $X$.

Q.E.D.

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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, $$ U = \bigcup_{\alpha\in I}\bigcap_{i\in J_\alpha} S_{\alpha, i} $$ where each $J_\alpha$ 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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  • $\begingroup$ Do you mean to say $$U=\cup_{\alpha \in I} \cap_{i \in J_{\alpha}}$$ and not $$U=\cup_{\alpha \in I} \cap_{i \in J_{i}}$$? $\endgroup$
    – Kam
    Dec 12, 2018 at 11:10
  • $\begingroup$ @Kam That appears to have been a typo, thanks. $\endgroup$
    – Math1000
    Dec 13, 2018 at 14:58

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