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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.

  • 3
    How do you define the topology on $X\times Y$?2012-11-29
  • 4
    ProofWiki: [Projection from Product Topology is Open](http://www.proofwiki.org/wiki/Projection_from_Product_Topology_is_Open)2013-04-11

3 Answers 3

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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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    Hi Jesko, nice answer, thanks. It is very helpful for a question I just asked.2013-07-05
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    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~2013-07-05
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    @WishingFish: See my edit. Hope this helps!2013-07-05
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    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.2013-07-05
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    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.2014-05-22
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    That's wrong, $f(U\cap V)\subset f(U)\cap f(V)$ generally. See the very page you have linked.2015-07-05
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    @GFR: Indeed, that was a serious blunder. Should be fixed now.2015-07-05
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    Think so, I have removed my down vote. You may want to edit your last line as well.2015-07-05
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    @GFR: True & done. Thanks for the comment btw, I realize now that tomasz already pointed it out before but I completely overlooked that in his comment.2015-07-05
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    I don't see how you can claim all open sets in the product topology are of the form VxW for V open in X and W open in Y. If I take another open set JxK, for J open in X and K open in Y, then the union of these two sets is open in the product topology but is not necessarily able to be formed via any crossproduct of open sets in X and Y. @JeskoHüttenhain2018-03-01
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    @H_1317: read the edit ;-)2018-03-01
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    So, bear with me, sorry -- If X x Y is R^2, then I can take two non intersecting rectangles/circles as my open sets in R^2 and take their union to create another open set in my product topology. This open set you call U, which represents any open set in R^2. However, though you show for our projection we can take the union of the V_1 an V_2 in R to see the projection maps to an open set (thus completing the proof), it does not seem true that the union of my two original circles/rectangles can be represented in the form UxV for any U in R and V in R. Make sense? @JeskoHüttenhain2018-03-02
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    Dear @H_1317, not every open set in $X\times Y$ is of the form $U\times V$ with $U$ open in $X$ and $V$ open in $Y$, that is correct. But please note that I *never* claimed that, and do not understand your point. I can only refer you *again* to the edit where the reduction to the completely general case is done.2018-03-02
19

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.

3

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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    Do you mean to say $$U=\cup_{\alpha \in I} \cap_{i \in J_{\alpha}}$$ and not $$U=\cup_{\alpha \in I} \cap_{i \in J_{i}}$$?2018-12-12
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    @Kam That appears to have been a typo, thanks.2018-12-13