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Given is $X:=[0,1]\times[0,1]$ with the euclidian topology. Let $\sim$ be the equivalence relation given by $(s,t)\sim(s',t')$ if $s = s'$ and $t = t'$, if $s = s' = 0$, if $s=s'$, $t = 1$ and $t'=0$, or if $s = s'$, $t = 0$ and $t' = 1$.

I need to show that $X/\sim$ with the quotient topology and the disk $D^2:=\{(x,y):x^2 + y^2 \leq 1\}$ with the euclidian topology are homeomorphic.

$X/\sim$ and $D^2:=\{(x,y):x^2 + y^2 \leq 1\}$ with their respective topologies are homeomorphic if there exists a homeomorphism between them. I know that a function $f:X\to{Y}$ is called a homeomorphism if $f$ is bijective and $f$ and $f^{-1}:Y\to{X}$ are both continuous. I just don't know how to find this function.

Q: How do I show that $X/\sim$ with the quotient topology and the disk $D^2:=\{(x,y):x^2 + y^2 \leq 1\}$ with the euclidian topology are homeomorphic?

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    $\begingroup$ Your third condition on $\sim$ includes $s=1'$. Is this supposed to be $s=1$ or $s=s'$? $\endgroup$ – Aweygan Jun 18 '17 at 12:57
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Let define $f\colon X\rightarrow D^2$ by $f(s,t):=(s\cos(2i\pi t),s\sin(2i\pi t))$, then $f$ is surjective and continuous. Furthermore, notice that: $$f(0,t)=f(0,t')\textrm{ and }f(s,0)=f(s,1).$$ Therefore, $f$ gives rise through the quotient to a bijective and continuous map denoted by $\overline{f}\colon X/\sim\rightarrow D^2$. Besides, since $X/\sim$ is compact, $\overline{f}$ is a homemorphism. Whence the result.

Remark. The set $X/\sim$ is compact since it is the direct image of $X$ by $X\twoheadrightarrow X/\sim$ which is continuous.

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The other answer works perfectly. Also consider that in general when trying to prove two spaces $X$ and $Y$ are homeomorphic especially when there is equivalence relation is to use the theorem that a continuous bijection from a compact space into a Hausdorff space is a homemorphism. The reason why it can't be applied directly is because without the equivalence relation the map you define might fail to be injective, the equivalence relation, assuming it's reasonable, will "squish" all the points of the function which fail to be injective into one point, and so the map from the quotient space to the target space IS bijective.

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