Jtdylktuy

Let $X$ and $Y$ be (any) topological spaces. Show that the projection

$pi_1$ : $Xtimes Yto X$

is an open map.

Let $Usubseteq Xtimes Y$ be open. Then, by definition of the product topology, $U$ is a union of finite intersections of sets of the form $pi_X^{-1}(V)=Vtimes Y$ and $pi_Y^{-1}(W)=Xtimes W$ for $Vsubseteq X$ and $Wsubseteq Y$ open. This means (in this case) that we may without loss of generality assume $U=Vtimes W$. Now, clearly, $pi_X(U)=V$ is open.

Edit I will explain why I assume $U=Vtimes W$. In general, we know that $U=bigcup_{iin I} bigcap_{jin J_i} V_{ij}times W_{ij}$ 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_1times W_1)cap (V_2times W_2) &= { (v,w) mid vin V_1, vin V_2, win W_1, win W_2 } \&= (V_1cap V_2)times (W_1cap W_2)
end{align*}
and this generalizes to arbitrary finite intersections. Now, we have
begin{align*}
pi_X(U)&=pi_Xleft(bigcup_{iin I}~ bigcap_{jin J_i} V_{ij}times W_{ij}right)
=bigcup_{iin I}~ pi_Xleft(left(bigcap_{jin J_i} V_{ij}right)times left(bigcap_{jin J_i} W_{ij}right)right)
= bigcup_{iin I}~ bigcap_{jin J_i} V_{ij} =: V
end{align*}
and $Vsubseteq X$ 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.

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.

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 $Xtimes Y$. Then $U$ is a union of finite intersections of elements of
$$ mathcal S = left{pi_1^{-1}(A) : Atext{ open in } Xright} cup left{pi_2^{-1}(B) : B text{ open in } Yright},$$
that is,
$$ U = bigcup_{alphain I}bigcap_{iin J_alpha} S_{alpha, i} $$
where each $J_alpha$ is finite and each $S_{alpha, i}$ is in $mathcal S$. We can write each $S_{alpha,i}=pi_1^{-1}(V_{alpha,i})cappi_2^{-1}(W_{alpha,i})$, 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 $$ pi_1^{-1}(V_{alpha,i})=V_{alpha,i}times Y text{ and } pi_2^{-1}(W_{alpha,i})=Xtimes W_{alpha,i}$$
it follows that
$$pi_1^{-1}(V_{alpha,i})cappi_2^{-1}(W_{alpha,i}) = (V_{alpha,i}times Y)cap (Xtimes W_{alpha,i}) = V_{alpha,i}times W_{alpha,i}.$$
Letting $V_alpha=bigcap_{iin J_i} V_{alpha,i}$ and $W_alpha = bigcap_{iin J_i}W_{alpha,i}$, we have
$$U = bigcup_{alphain I}bigcap_{iin J_i} V_{alpha,i}cap W_{alpha,i} = bigcup_{alphain I}V_alphatimes W_alpha,$$
where each $V_alpha$ is open in $X$ and each $W_alpha$ is open in $Y$. It follows that
$$ pi_1(U) = pi_1left(bigcup_{alphain I}V_alphatimes Walpha right) = bigcup_{alphain I}pi_1(V_alphatimes W_alpha) = bigcup_{alphain I'}V_alpha $$
(where $I' = {alpha in I : W_alphanevarnothing}$) is open in $X$. We conclude that $pi_1$ is an open map.