# Projection of topological space $X\times Y$ is an open map

I must prove that the projections $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$ are open, that is, they take open sets to open images. Since a basis element of $X\times Y$ is of the form $U\times V$ where $U$ is open in $X$ and $V$ is open in $Y$. However, the collection of basis doesn't form a topology, since the intersection of two rectangles won't be a rectangle. Therefore, in order to assume an element of the set $X\times Y$, I need to take an union of basis elements, that is, $W = \cup_{i\in I} U_i\times V_i$.Therefore, I'm talking about the projection from an union to a set.

I know I could see an element of the product topology as an union of finite intersections, as seen here, but this is too complicated and my professor explicitly told us to search if the projection of a union of the form $W$ is the union of the projections, then that would be easy. Could somebody help me?

• The intersection of two rectangles is a rectangle (you deny that). The union of two (or more) rectangles is not necessarily a rectangle. – drhab Sep 9 '16 at 22:35
• @drhab ops, I meant the union, thanks – Guerlando OCs Sep 9 '16 at 22:36

## 1 Answer

If $W$ is open in $X\times Y$ then it can be written as a union of rectangles: $W=\bigcup_{i\in I}(U_i\times V_i)$ where $U_i$ is open in $X$ for every $i\in I$ and $V_i$ is open in $Y$ for every $i\in I$.

Consequently: $$\pi_1(W)=\pi_1\left(\bigcup_{i\in I}(U_i\times V_i)\right)=\bigcup_{i\in I}\pi_1(U_i\times V_i)=\bigcup_{i\in I}U_i$$ and:$$\pi_2(W)=\pi_2\left(\bigcup_{i\in I}(U_i\times V_i)\right)=\bigcup_{i\in I}\pi_2(U_i\times V_i)=\bigcup_{i\in I}V_i$$

Both sets are unions of open sets hence are open.

• but why the projection of the union is the union of projections? That's the main question :( – Guerlando OCs Sep 9 '16 at 22:55
• In general $f(\bigcup_{i\in I}A_i)=\bigcup_{i\in I}f(A_i)$. Observe that for a fixed $j\in I$ we have $A_j\subseteq\bigcup_{i\in I}A_i$ and consequently $f(A_j)\subseteq f(\bigcup_{i\in I}A_i)$. Conversely if $y\in f(\bigcup_{i\in I}A_i)$ then $y=f(x)$ for some $x\in\bigcup_{i\in I}A_i$. Then $x\in A_j$ for some fixed $j\in I$ so that $y=f(x)\in f(A_j)\subseteq \bigcup_{i\in I}f(A_i)$. – drhab Sep 9 '16 at 23:06
• are yoy saying that the image of an union is always the union of images? I've never heard of such a theorem or property, but I understood your proof. This seems important but I've never seen it – Guerlando OCs Sep 9 '16 at 23:44
• That is indeed what I am saying. I think that quite often it is not presented as theorem but is given as an exercise. As you can see there is no real deepness in the proof. This might explain that you have never seen it. – drhab Sep 10 '16 at 6:58