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Let $X=S^1 \times I$ be a cylinder, where $S^1$ is the 1-dimensional circle. If we glue the "bottom" boundary $S^1 \times 0$ and the "top" boundary $S^1\times 1$ by a homeomorphism sending $x\times 0$ to $x\times 1$, the resulting manifold is homeomorphic to a torus.

If we chose another homeomorphism to glue boundaries, can we get a manifold that is not homeomorphic to a torus? (I only consider oriented manifolds so Klein bottle are omitted.)

Or no matter what homeomorphism we choose, is the resulting manifold is homeomorphic to a torus?

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Consider the orientations on the circles $S^1 \times 0$ and $S^1 \times 1$ coming from a choice of orientation on $S^1$. If the gluing map $f : S^1 \times 1 \to S^1 \times 0$ preserves orientation, the quotient is homeomorphic to a torus. If not, it is homeomorphic to a Klein bottle. The proof requires one to know that two self-homeomorphisms of $S^1$ are isotopic if they both preserve orientation or if they both reverse orientation. The homeomorphism between the quotient spaces of $S^1 \times [0,1]$ may then be constructed by "absorbing" the isotopy into the product structure on $S^1 \times [0,1]$.

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