How to prove this map is continuous

Question

Let $Y=\mathbb{R}^2 \setminus\left\{\begin{bmatrix}0\\ 0 \end{bmatrix}\right\}$ and $I=\left[0,1\right]$, both with the subspace topologies of the Euclidean ones.

Define a map $F:Y\times I\to Y$ by $$F\left(z,\,i\right):=\frac{z}{\left|z\right|}\left(\left(\left|z\right|-1\right)i+1\right)\forall\left(z,\,i\right)\in Y\times I$$

What is the easiest way to prove that $F$ is continuous?

I have tried taking an arbitrary open ball $B$ in $Y$ and showing that $F^{-1}(B)\in Open(Y\times I)$, but I'm not sure just how to do that. I tried to picture what $F^{-1}(B)$ looks like. I know that a point on some ray can only come from another on the same ray, but I don't think it is in general true that $F^{-1}(B)=V\times W$ where $V\in \operatorname{Open}(Y)$ and $W\in \operatorname{Open}(I)$.

Since the domain doesn't include the origin, isn't this function just the composition of a bunch of continuous functions, and thus continuous? — Ashwin Iyengar, Dec 23 '14 at 22:14

score 3 · Accepted Answer · answered Dec 23 '14 at 22:15

The easiest way to prove it "from scratch" is to use the $\delta$-$\epsilon$ definition. That is:

Let $z,i$ be arbitrary. Prove that for every $\epsilon > 0$, there exists an open neighborhood $U$ of $(z,i)$ such that if $(z',i') \in U$, then $\|F(z,i) -F(z,i)\| < \epsilon$.

Alternatively, it suffices note that $F$ is the composition of several continuous functions.

How to prove this map is continuous

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