Help with my proof of : Show that the projection map $\pi: X \times Y \to X$ on the product topology is an open map.

Question

I saw some other answers on here that seemed really involved. Munkres defined the product topology on $X \times Y$ as the topology having as basis the collection of all sets of the form $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$.

A map $f: X \to Y$ is open if for every open set $U$ of $X$, $f(U)$ is open in $Y$.

So, if $U \times V$ is open in $X \times Y$, then $U$ is open in $X$. But $\pi(U \times V)=U$ which is open in $X$.

Is there something wrong with this?

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I also have: Let $U \times V$ be an open set containing $(x,y)$ and let $A\times B \subset U\times V$ be a basis element containing $(x,y)$. So, $$(x,y)\in A\times B \subset U\times V$$ $$\implies \pi((x,y)) \in \pi(A\times B)\subset \pi(U\times V)$$

But I am not sure of this.

Answer 1

Remember the product sets are just a basis, you need to do a teensy bit more work since not all open sets are product sets. Let $(x,y)\in U\times V\subseteq W\subseteq X\times Y$ be a point with a neighborhood $W$. Then

$$\pi(W)\supseteq \pi(U\times V) = U\ni \pi(x,y)=x$$

So the image of every point in an open set has around it an open set, hence $\pi(W)$ is open.

Answer 2

Hint:

Your first argument goes wrong because an open subset of $X \times Y$ can be much more complicated than the product $U \times Y$ of open subsets of $X$ and $Y$. E.g., take $X = Y= \Bbb{R}$ and think about the set of points $(x, y) \in \Bbb{R}\times\Bbb{R}$ with $x > y$.

You can answer the question thinking just about open sets rather than elements and their neighbourhoods: by the definition of the product topology, an open subset $P$ of $X \times Y$ is a union of sets of the form $U \times V$ where $U$ and $V$ are non-empty open subsets of $X$ and $Y$ respectively. Say $P = \bigcup_{i\in I} (U_i \times V_i)$ where the $U_i$ are non-empty and open in $X$ and the $V_i$ are non-empty and open in $Y$. What can you say about $\pi(P)$?

Answer:

$\pi(P) = \bigcup_{i\in I} U_i$. So it's the union of a family of open subsets of $X$ and hence is open in $X$.