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Problem. Let $T=S^1\times S^1$, where $S^1=\{z\in\mathbb{C}:|z|=1\}$. Prove the quotient space of $T$ by the equivalence relation $(z,w)\sim(\bar{z},-w)$ is homeomorphic to the Klein bottle.

Theorem I. Let $X$ be compact and $Y$ be Hausdorff. If $f:X\rightarrow Y$ is continuous and surjective with the point inverses in $Y$ under $f$ being the equivalence relation $\sim$ on $X$, then $X/\sim$ is homeomorphic to $Y$.

Attempt. The Klein bottle is $K=I \times I/[(t,0)\sim(t,1),(0,t)\sim(1,1-t)]$. I want to use (I). The equivalence relation that we must factor by represets a reflection of each point through the origin (right?). I'm having trouble finding a continuous function that has preimages make up the equivalence class.

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  • $\begingroup$ It depends on how much topology you know. For instance, do you already know the classification of surfaces? $\endgroup$ – Moishe Kohan Oct 16 '19 at 1:47
  • $\begingroup$ I have not yet learned classiciation of surfaces. $\endgroup$ – Saru Oct 16 '19 at 1:57
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Let me rewrite your notation for $K$ to avoid a modest clash of variables: $$K = I \times I \, / \, [(s,0) \sim (s,1), (0,t) \sim (1,1-t)] $$ And now I want to switch the notation for $T = S^1 \times S^1$ to something which more closely matches that notation for $K$, namely $$T = I \times I \, / \, [(s,0) \sim (s,1), (0,t) \sim (1,t)] $$ The relation between this notation and the notation $(z,w) \in S^1 \times S^1$ is $z = e^{2 \pi i s}$ and $w = e^{2 \pi i t}$.

When you now convert the equivalence relation $\sim$ on $T$ from $(z,w)$ notation to $(s,t)$ notation, you get $$(s,t) \sim \begin{cases} (1-s,t+1/2) &\text{if $0 \le t \le 1/2$} \\ (1-s,t-1/2) &\text{if $1/2 \le t \le 1$} \end{cases} $$ Since every point in the bottom half $[0,1] \times [0,1/2]$ is equivalent to a point in the top half $[0,1] \times [1/2,1]$, and since those two sets are compact, we can rewrite the quotient space $T / \! \sim$ by discarding all but the bottom half, to get the following gluing pattern: $$T / \!\sim \, = \, [0,1] \times [0,1/2] \, / \, [(s,0) \sim (1-s,1/2), (0,t) \sim (1,t)] $$ And now you can exactly match this gluing pattern with the gluing pattern for $K$ by using the map $[0,1] \times [0,1/2] \mapsto [0,1] \times [0,1]$ given by the formula $(s,t) \mapsto (2t,s)$.

And, by the way, this proof does indeed use Theorem I, to justify that the quotient is unchanged up to homeomorphism after throwing away all but the bottom half.

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