# Proving Quotient Space of Torus Homeomorphic to Klein Bottle

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.

• It depends on how much topology you know. For instance, do you already know the classification of surfaces? – Moishe Kohan Oct 16 '19 at 1:47
• I have not yet learned classiciation of surfaces. – Saru Oct 16 '19 at 1:57

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 $$(s,t) \mapsto (2t,s)$$.