I am trying to determine the fundamental group of the following parametrized surface: $$ X(r,\theta) = (r\cos\theta,r\sin\theta, ln(r^2)),$$ where $r\in(0,+\infty)$ and $\theta \in [0,2\pi)$. It can be thought of as the surface obtained when revolving the one-variable logarithm around the z-axis. Here's my problem.
On the one hand, I think that this surface is homeomorphic to the punctured plane. Hence, its fundamental group should be $\mathbb{Z}$.
On the other hand, I have come up with an apparently convincing proof that the one-variable logarithm is a strong deformation retract of this surface. Since the one-variable logartihm is just a line, it is homeomorphic to the real line $\mathbb{R}$. Hence, the fundamental group should be trivial.
This is indeed a contradiction. And I cannot seem to find the mistake I am making. Now I will include the strong deformation retract that I have come up with, which is likely to be mistaken. I claim $R$ is a retract of $X$:
$$ R = \{ (r,0,\log(r^2)) \,|\, r \in (0,+\infty) \} $$ $$ H: X \times [0,1] \longrightarrow X$$ $$ H( X(r,\theta) , t ) = ( r\cos(1-t) \theta , r \sin(1-t)\theta , \log(r^2)).$$
It seems to satisfy the conditions of a strong deformation retract, namely $ H(p,0)=p \,\,\forall \,\,p \in X $, $ H(p,t)=p \,\,\forall p \,\, \in R $ and $ H(p,1)\in R \,\,\forall \,\,p \in X $. Here's a picture of X (blue) and R (black):
The first approach is correct. This surface is homeomorphic to the punctured plane (by projection on the first two coordinates, for example). Your $H$ is not a continuous map on the surface $X$. This is because the map $X(r,\theta)\mapsto(r,\theta)$ is not continuous, even though it is bijective. For example, the sequence $(X(1,2\pi-1/n))_n$ converges to $X(1,0)$, but $(\cos((1-t)(2\pi-1/n)),\sin((1-t)(2\pi-1/n)),0)\rightarrow(\cos(2\pi(1-t)),\sin(2\pi(1-t)),0)$ and this does not equal $(1,0,0)$ for $0<t<1$.