Let $T^2=S^1\times S^1$ be the torus and $p\in T^2$. Show that the punctured torus $T^2-\{p\}$ has the figure eight $S^1\vee S^1$ as a deformation retract.
The torus $T^2$ is homeomorphic to the quotient space: $$I\times I/\sim$$ where $I=[-1,1]$ and $\sim$ is the equivalence relation generated by $(1,t)\sim (-1,t)$ and $(s,-1)\sim (s,1)$. Also, $S^1\vee S^1$ is homeomorphic to $\partial (I\times I)/\sim$.
Now, it is easy to show that the family of maps $\{f_\lambda\}$ given by: $$f_\lambda(s,t)=\left\{\begin{array}{lll} \lambda(\frac{s}{|s},\frac{t}{|t|})+(1-\lambda)(s,t), &\text{if}& |s|\geq |t|\\ \lambda(\frac{s}{|t|},\frac{t}{|t|})+(1-\lambda)(s,t), &\text{if}& |t|\geq |s| \end{array}\right.$$ defines a deformation retraction of $I\times I-\{(0,0)\}$ onto $\partial (I\times I)$.
My question is:
- How can I prove that the family $\{f_\lambda\}$ induces a retraction given by: $$\begin{array}{rcll} \hat f_\lambda:&T^2-\{p\}\cong (I\times I-\{(0,0)\})/\sim &\longrightarrow & S^1\vee S^1\cong \partial(I\times I)/\sim\\ &[(s,t)]&\longmapsto &[f_\lambda(s,t)] \end{array}$$ ?
- Is there some property that ensures that it is a retraction between the quotient spaces?