A map $f:X\to Y$ is said to be an open map if for every open set $U$ of $X$, the set $f(U)$ is open set in $Y$. Show that $\pi_{1}:X\times Y\to X$ and $\pi_{2}:X\times Y\to Y$ are open maps.
I rephrase this exercise to my taste.
If $(X,\mathcal{T}_{X})$, $(Y,\mathcal{T}_{Y})$, the set $X\times Y$ equipped with product topology, denoted by $\mathcal{T}_{p}$, then $\pi_{1}:X\times Y\to X$ and $\pi_{2}:X\times Y\to Y$ are open maps.
My attempt: let $U\subseteq X\times Y$ and $U\in \mathcal{T}_{p}$. We need to show $\pi_{1}(U)\in \mathcal{T}_{p}$. Since $U\in \mathcal{T}_{p}$, we have $U= \bigcup_{i \in I}(U_i\times V_i)$, where $U_i\times V_i \in \mathcal{B}, \forall i\in I$. We claim $\pi_{1}(U)=\cup_{i\in I} U_i$. Let $x\in \pi_{1}(U)=\{ u|u\times v\in U\}$. Take $y\in Y$ such that $x\times y\in U$. since $x\times y\in U, \exists j\in I$ such that $(x,y)\in U_j\times V_j$. Which means $x\in U_j\subseteq \cup_{i \in I}U_i$. So $x\in \cup_{i \in I}U_i$. Thus $\pi_{1}(U)\subseteq \cup_{i\in I} U_i$.
Conversely, let $x\in \cup_{i\in I} U_i$. Then $\exists j\in I$ such that $x\in U_j$. Take $y\in Y$ such that $y\in V_j$. So $(x,y)\in U_j\times V_j \subseteq \bigcup_{i\in I}(U_i \times V_i)= U$. So $(x,y)\in U$. Thus $x\in \pi_{1}(U)$. Hence $\pi_{1}(U)\supseteq \cup_{i\in I} U_i$. So $\pi_1(U)$ is arbitrary union of open sets in $X$. $\pi_1(U)$ is open in $X$. A similar proof show $\pi_2$ is open. Is this proof correct?
Approach(2)
let $U\subseteq X\times Y$ and $U\in \mathcal{T}_{p}$. $\forall (x,y)\in U, \exists R_{(x,y)}\times S_{(x,y)}\in \mathcal{B}$ such that $(x,y)\in R_{(x,y)}\times S_{(x,y)}\subseteq U$. Since $U\subseteq X\times Y$, $U=M\times N$, where $M=\pi_1 (U)$ and $N=\pi_2 (U)$. Thus $\cup_{(x,y) \in U} R_{(x,y)}= \pi_1 (U)$ and $\cup_{(x,y) \in U} S_{(x,y)}= \pi_2 (U)$. Our desired result. Is this proof correct?
