Is the mapping $\mathbb{R}^2 \to \mathbb{R} ^2$ given by $(\xi_1,\xi_2) \to (\xi_1,0) $ an open mapping ?Yes/No

Question

Is the mapping $f:\mathbb{R}^2 \to \mathbb{R} ^2$ given by $(\xi_1,\xi_2) \to (\xi_1,0) $ an open mapping ?Yes/No

My attempt : yes ,I think it is open because collection of elements of the form $(\xi_1,0)$ is open in $\mathbb{R}^2$

Let $U\subseteq\Bbb R^2$ be open. Suppose $z\in f(U)$. Then there is $(\xi_1,0)\in U$ with $\xi_1=z$. As $U$ is open, there exists $r>0$ such that $(\xi_1-r,\xi_1+r)\subseteq U$. Conclude that $(z-r,z+r)\subseteq f(U)$.

Answer 1

Hint: Is $]z-r, z+r[\times\{0\}$ an open subset of $\mathbb R^2$ ?

Answer 2

Then there is $\xi_1 \in U$ with $\xi_1 = z$.

No. $z \in f(U) \subseteq \mathbb{R}^2$, so $z$ is an ordered pair, not a single number $\xi_1$.

Conclude that $(z - r, z + r) \subseteq f(U)$.

What does this mean? $(z - r, z + r) \in \mathbb{R}^2$ is an ordered pair of numbers, so it doesn't make sense to say it's a subset of $f(U)$.

Do you mean $(z - r, z + r) \in f(U)$? If so, you should prove this by showing there exists $a, b$ such that $f(a, b) = (z - r, z + r)$.