Homeomorphism between topological spaces

Question

I need to check the following exercise:

Let $X=[0,1] \times [0,1]$ and let $\sim$ be a relation defined by $(x,0) \sim (x,1)$. Show that $X/ \sim$ is homeomorphic to the cylinder $S^1 \times [0,1]$.

Solution:

Consider the function $f: X \rightarrow [0,1] \times S^1$ defined as $f(\theta, t)=(t(t-1), \cos(2 \pi \theta), \sin (2 \pi \theta))$. Since $f$ is conituous, bijective and $f(0,0)=(0,1,0)=f(0,1)$ we have that $[f]:X/ \sim \longrightarrow [0,1] \times S^1$ is continuos too.

Moreover, since $f^{-1}:[0,1] \times S^1 \rightarrow X$ defined by $f^{-1}(x,y,z)=\left ( \sqrt{x-1}, \frac{\arccos y}{2 \pi}, \frac{\arcsin z}{2 \pi}, \right )$ is continuos on its and bijective on its domain, we have that $[f]^{-1}$ is continuous, hence homeomorphism.

Does it work?

Answer 1

Your $f$ is not injective. For instance

$$ f(0, 1/4) = (1, 0, 3/16) $$ and $$ f(0, 3/4) = (1, 0, 3/16). $$

I think you might want to look at the order of the two factors.

NB

The question has been changed (to yet another non-injective function) after I provided this answer. The function, when I answered, was $$ f: X \rightarrow S^1 \times [0,1]: (\theta, t) \mapsto (\cos(2 \pi \theta), \sin (2 \pi \theta), t(t-1)). $$

To critique this new answer, which says (I quote this because I anticipate it being changed again and again and again)

$f: X \rightarrow [0,1] \times S^1$ defined as $f(\theta, t)=(t(t-1), \cos(2 \pi \theta), \sin (2 \pi \theta))$. Since $f$ is conituous, bijective and $f(0,0)=(0,1,0)=f(0,1)$ we have that $[f]:X/ \sim \longrightarrow [0,1] \times S^1$

It's worth nothing that when $t = 0.5$, the first output component, $t(t-1) = 0.5 (-0.5) = -0.25$ is not actually in the interval $[0, 1]. So this answer's still wrong.