I will only quote the conditions that are relevant to my problem.
Let $f: A \subseteq \mathbb{R}^{n+k} \to \mathbb{R}^n$ where $A$ is open be a $C^r$ map with $f(a,b)=0$.
In the proof, a function $F: A \to \mathbb{R}^{k+n}$ is then defined by $F(x,y) = (x, f(x,y))$ where $x$ is a vector of length $k$, $y$ a vector of length $n$.
After some argumentation using other conditions, it is proven that $DF(a,b)$ is invertible and $F(a,b) = (a,0)$.
The inverse function theorem is then used to conclude that there are open sets $V$ in $\mathbb{R}^k$ and $U$ in $\mathbb{R}^n$ and $W$ in $\mathbb{R}^{k+n}$ such that
$$F: U \times V \to W$$ is a bijection with inverse of class $C^r$ and $a \in U, b \in V, (a,0) \in W$.
I can see that the inverse function theorem gives an open set $O \subseteq \mathbb{R}^{k+n}$ in the domain on which $F$ is 1-1 on its image, but why can we write $O = U \times V$ for such open sets?
I can see we have $U \times V \subseteq O$ for some open sets $U$ and $V$ using the properties of the product topology.