# Product maps are open maps

I am trying understanding product maps better. Suppose that we have open maps$$f_1:X_1\rightarrow Y_1, f_2:X_2\rightarrow Y_2$$ Then, if $U_i\subseteq X_i \Rightarrow f_i(U_i)\subseteq Y_i$ for $i=1,2$.

Now consider, $$f_1\times f_2:(X_1\times X_2)\rightarrow(Y_1\times Y_2)$$ I believe that this should also be an open map.

Let $U$ be an open subset of $X_1\times X_2$, then $U$ can be expressed as $U_1\times U_2$ where $U_1\subseteq X_1$ and $U_2\subseteq X_2$. I want to show that $(f_1\times f_2)(U)$ is an open subset of $Y_1\times Y_2$. If $x=(x_1,x_2)\in U$, then $(f(x_1),f(x_2))\in (f_1\times f_2)(U)$. Also, $f_1(U_1)$ is open in $Y_1$ and $f_2(U_2)$ is open in $Y_2$. Here is where I am stuck and having difficulty continuing.

Hints are welcome, please no full solutions. I want to work through this. Thank you for your help.

• Open sets are unions of products of basic open sets, not products. For example, an open disk in the plane is not a product of two open intervals. – Alex Provost Oct 23 '17 at 0:54
• Open sets needn’t be products of open sets. But they are unions of such products since products of open sets form a basis for the product space. – MPW Oct 23 '17 at 0:54
• Instead, should it be: If $U$ is open in $X_1\times X_2$, then there are $U_1,U_2$ such that $U_1\times U_2\subseteq U$? – Joe Oct 23 '17 at 0:55
• Prove the easy and useful theorem that if B is a base for X and for all U in B, f(U) is open, then f is open. – William Elliot Oct 23 '17 at 1:17
• If $U$ is open in $X_1\times X_2$ then $U=\cup G$ where each $g\in G$ is $U_{1,g}\times U_{2,g}$ for some open $U_{i,g}$ of $X_i$..... So $(f_1\times f_2)(U)=$ $\cup_{g\in G} (f_1\times f_2)(g)=$ $\cup_{g\in G}[\;(f_1\times f_2)(U_{1,g}\times U_{2,g})\;].$ ... To solve your problem, answer this : What is the definition of $(f_1\times f_2)(U_{1,g}\times U_{2,g})$ ? – DanielWainfleet Oct 23 '17 at 16:22

$U \subseteq X \times Y$ is open iff for every $(x,y) \in U$ there exist $U_1 = U_1(x) \subseteq X$ open, and $U_2 = U_2(x) \subseteq Y$ with $x \in U_1, y \in U_2$ and $U_1 \times U_2 \subseteq U$.
Now check that $\pi_1[U] = \bigcup_{(x,y) \in U} U_1(x)$, which is a union of open sets and thus open.
• This helps a lot. I was under the "false" impression that if $U$ was open in $X_1\times X_2$, then $U=U_1\times U_2$. However, by definition, the product topology is generated by the basis $\{U_1\times U_2 : U_1\subseteq X_1, U_2\subseteq X_2\}$. – Joe Oct 25 '17 at 12:27