Suppose $\mathbb{R}^2$ and $\mathbb{R}$ are topological spaces with standard topology. Let $f(x,y) = x + y^2$ .
How do I show that for any open set $U$ in $\mathbb{R}^2$, $f(U)$ is an open set in $\mathbb{R}$?
This looks similar to this theorem: Let $X$ and $Y$ be two topological spaces, and $\pi: X \times Y \to X$ , $\pi(x,y) = x$. If $O$ is open in $X \times Y$, then $\pi(O)$ is open in $X$.
Let $\newcommand{\R}{\mathbb R}t\in f(U)$ be given. Find $(a,b)\in U$ so that $f(a,b)=t$. Since $U$ is open, it is possible to find an open ball with radius $2r$: $$ B_{2r}(a,b)\subset U $$ and within that open ball we have an open square: $$ Q:=I_r(a)\times I_r(b)\subset B_{2r}(a,b) $$ where the $I_r(x)=(x-r,x+r)$ denotes an open interval around $x\in\R$. Now define $f_x(x,y)=x^2$ and $\varphi(x,y)=y$. With this the claim is that we have $$ f(Q)=f_x(Q)+\varphi(Q) $$ where a sum of two sets $A+B$ denotes the set containing all sums of pairs of elements $a+b$ with $a\in A$ and $b\in B$. Clearly $f(Q)\subseteq f_x(Q)+\varphi(Q)$ since $k\in f(Q)$ has $(q,s)\in Q$ that maps to $k=f(q,s)=q^2+s=f_x(q,s)+\varphi(q,s)$.
On the other hand, given $k\in f_x(Q)+\varphi(Q)$ we have $k=r+s$ with $r\in f_x(Q)$ and $s\in \varphi(Q)$, but then $f_x^{-1}(r)\cap Q$ are points $(q,y)$ in $Q$ so that $q^2=r$ and $y$ can be chosen arbitrarily in $I_r(b)$, whereas $\varphi^{-1}(s)\cap Q$ are points $(x,s)$ in $Q$ where $x$ can be chosen arbitratily in $I_r(a)$. We must have $q\in I_r(a)$ and $s\in I_r(b)$ so that $(q,s)$ is an element of $Q$. This was the step where the definition of the square $Q$ in $\R^2$ became crucial! Thus $k=f(q,s)$ showing that $k\in f(Q)$.
What is left to show is that when you add two sets $A,B\in\R$ and one of them is an open set, then the sum $A+B$ is open as well. Then it follows that $f(Q)=f_x(Q)+\varphi(Q)$ is a way to write $f(Q)$ as a sum of sets in $\R$ where $\varphi(Q)$ will be open by the theorem that you qouted, so it follows that $f(Q)$ must be an open set containing $t$ and therefore we can find an open interval around any given $t\in f(U)$. It follows that $f(U)$ is open.
I left to you to show that $A+B$ will be open whenever one of the summands are...
There is a much easier way to do this, using exactly the observation by the OP that the projection map $\pi \colon X \times Y \to X$ is open.
The map $f$ is the composition of the homeomorphism $(x,y) \mapsto (x + y^2, y)$ (which has as inverse $(x,y) \mapsto (x - y^2, y)$) and the projection $\pi \colon {\mathbb R}^2 \to {\mathbb R}$ on the first coordinate. Since the composition of open maps is open, $f$ is open.
