Since in my previous post I couldn't find the correct solution to the following problem, I repost it, and since I'm not an expert in topology I kindly ask if somebody can give me the correct answer.
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 $S^1 \times [0,1]$.
My goal is to get a solution without using compactness, writing a continuos and bijective function $f: X \rightarrow Y$, where $Y$ is the cylinder, constant on the equivalente classes of $X/\sim$. This implies that the function $[f]: X/\sim \rightarrow Y$ is continuos and bijective too. Since the composition $f= [f] \circ \pi$ is commutative (where $\pi$ is the natural projection) I'll have to show that also $f^{-1}$ is continuos, and then $[f]^{-1}$ is continuos too. Hence $[f]$ is an homeomorphism.
Solution:
Consider the function $f: X \rightarrow S^1 \times [0,1]$ defined as $f(\theta, t)=(\cos(2 \pi \theta), \sin (2 \pi \theta), t(t-1))$. Since $f$ is continuous, and $f(0,0)=(1,0,0)=f(0,1)$ we have that $[f]:X/ \sim \longrightarrow S^1 \times [0,1]$ is continuos too.
The problem is that $f$ is not injective, then is not invertible. Does it exist a suitable bijective function or it is impossible to claim that the two spaces are homeomorphic without using the compactness?
You can construct an explicit bijection. Your attempt is close, but $f(t, \theta)$ should read $f(\theta, t)$ and $t(t-1)$ should read $t$ (it's the trig functions that identify the case where $\theta = 0$ and the case where $\theta = 1$).
Geometrically, think of the cylinder $S^1 \times [0, 1]$ being obtained from the square $[0, 1]\times[0, 1]$ by rolling it up so the top edge meets the bottom edge. (The way the question is formulated also swaps the coordinates around: as if you'd stood the cylinder on its end.)