On the existence of a continuous bijection from a quotient space to the unit sphere $S^2$ There is a question from an old topology prelim that is somewhat giving me a hard time. Consider the cylinder $X= S^1 \times [-1,1]$. Now we define an equivalence relation $\sim$ as follows: For points $v,v' \in S^1$, we have $(v,-1) \sim (v',-1)$ and $(v,1) \sim (v',1)$. I am asked to show that the quotient space $X^{*}= S^1 \times [-1,1]/\sim$ is homeomorphic to the unit sphere $S^2$. The problem is I can't off the top of my head come up with a decent continuous bijection from the quotient space onto $S^2$. What might work here?  
Suppose I had some sort of continuous bijection $h: X^{*} \rightarrow S^2$. Now the quotient map $p: X \rightarrow X^{*}$ is continuous and surjective, and since $X$ is compact, so is $X^{*}$. We also know that $S^2$ being a topological manifold is Hausdorff. Recall that if there is a continuous bijection between the compact space $X^{*}$ (any compact space for that matter) and the Hausdorff space $S^2$ (or any Hausdorff space), then that continuous bijection is a homeomorphism. This is what I intended to do, but I still can't come up with such a continuous bijection. Also, perhaps I am a bit confused in trying to visualize the quotient space. I would really appreciate some input on this, and any ideas that may prove useful.
 A: Try to find for a point $((a,b),z)\in S^1\times [-1,1]$ a point $(x,y,z)$ in $S^2$ with the same $z$-coordinate and $x=\lambda a,\ y=\lambda b$ for a $\lambda$ which is a function in $z$.
Then show that the map $f:((a,b),z)\mapsto (x,y,z)$ is surjective and continuous, and that it induces a map $\tilde f:X^*\to S^2$. Is $\tilde f$ injective?
A: Here's an alternative approach which may or may not be what your topology prelim was asking for. The quotient space is gotten by crushing each end of the cylinder $S^1\times[-1,1]$ to a point. I claim this is a 2-manifold. This is clear away from the two crushed points. Let $p$ be the point which is gotten by crushing  $S^1\times\{1\}$. Let $U$ be the set of all equivalence classes of points $[(x,t)]$ in the quotient with $t>0$. I claim the map $f\colon D^2\to U$ given by $f(r,\theta)=[(\cos \theta,\sin\theta ,1-r)]$ is a homeomorphism. Here I am using polar coordinates in the unit disk $D^2$. (This step is probably just about equivalent in complexity to doing the problem a different way, but maybe it's slightly easier to see.) So $p$ has an open neighborhood homeomorphic to an open subset of $\mathbb R^2$. Symmetrically, the other quotient point is also a manifold point. So we know the quotient is a surface. Moreover, our argument shows that it is the union of two open disks and the circle at $t=0$, which is the disjoint union of a point and an open interval. So the Euler characteristic is $1-1+2=2$, and by the classification of surfaces, it must be a sphere.
A: The obvious projection map from $X$ (as a subset of $\mathbb R^3$) to the unit sphere in $\mathbb R^3$ is a continuous surjection. By the universal property of the quotient topology it descends to a map $f: X^* \to S^2$. Now show that $f$ has the desired property.
