Is the function $f:\mathbb D\to S^1\times I$ given in polar coordinates by $f(r,θ)=(θ,r)$ (or to be precise: $f(r\cos\theta,r\sin\theta)=((\cos\theta,\sin\theta),r)$) continuous?
How would one prove it?
My guess would be yes, since geometricly it is almost the identity, if looking on it with the right prespective. how would I prove it tho?
Thanks
P.S: This is standard notation in topology, but let me make it clear: $\mathbb D$ is the closed unit disk in $\mathbb R^2$, $S^1$ is it's boundary (the $\mathbb R^2$ sphere), and $I$ is the unit interval
P.S2:
Lemme explain my situation, and I hope you could help me use what I want properly. Let G:S1×I→X be homotopy between the constant y to a function g:S1→X i.e G is continuous with G(x,0)=y,G(x,1)=g(x). I want to define f:D→X by f(r,θ)=G(θ,r) and to claim f to be continuous. our discussion here should make f continuous in D{0,0} as a composition of continuous , and I hope that the fact that for r=0 it doesn't matter for G what θ is (it will always be y) will fix the continuity in (0,0). How could I show that?
The function $f$ is not defined at $(0,0)$. However it is continuous from $\mathbb{D}\setminus\{(0,0)\}$ to $S^1\times(0,1]$.
Let $X$ be a topological space (that for simplicity I will asume is a metric space with metric $d$) and $G\colon S^1\times[\,0,1\,]\to X$ continuous and such that $G(\theta,0)=y\in X$ for all $\theta\in S^1$. Define $f\colon\mathbb{D}\to X$ by $$ f(r,\theta)=\begin{cases} G(r,\theta) & r\ne0,\\ y & r=0.\end{cases} $$ Then $f$ is continuous. Continuity in $\mathbb{D}\setminus\{(0,0)\}$ is clear.
Let $\epsilon>0$ be given. Since $S^1\times[\,0,1\,]$ is compact, $G$ is uniformly continuous. There exists $r_\epsilon>0$ such that if $0\le r\le\epsilon$ then $d(G(r,\theta),y)\le\epsilon$ for all $\theta\in S^1$. Then also $d(f(r,\theta),y)\le\epsilon$.