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Can anyone suggest at least verbally how it is that a torus is constructed from two cells? (Total beginner to topology, been away from math as a whole for awhile...) Thanks for any insights!

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Does the diagram in page 32 of books.google.com/… help? –  Arturo Magidin Dec 13 '10 at 3:40
    
youtube.com/watch?v=0H5_h-RB0T8 –  Juan S Dec 13 '10 at 3:54
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@George Milly: Just cut the original square into four squares, by joining the midpoints of opposite sides; then you are not gluing any cell to itself in the process. (I'm not sure if by your title you mean getting the torus from 2 cells, or from any number of $2$-cells; if the former, then instead of breaking it into $4$, just cut along the diagonal to get two cells instead). –  Arturo Magidin Dec 13 '10 at 4:22
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@George Milly: In the video, think of those two blue circles as the 1-cells that were attached. Then you're not attaching the 2-cell to itself, you're attaching it to the 1-cells. –  Eric O. Korman Dec 13 '10 at 13:50
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A blind calculation that gives you the number of cells you need: Take a cell decomposition of $S^1$ (one 0-cell and one 1-cell). Then the torus $S^1\times S^1$ has as cells the products of the cells of the two $S^1$ factors, so we get one 0-cell, two 1-cells, and one 2-cell. –  Zach Conn Dec 13 '10 at 20:44

2 Answers 2

The torus can be made from a single 0-cell, two 1-cells and one 2-cell. The 1-cells are attached along their boundary to the 0-cell, making two circles identified at a point. Then the 2-cell is attached to this 1-skeleton, as illustrated in the comments above.

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The point of view offered by Jim is good because it can be generalized: if you take any surface with $n$ holes (say a connected sum of $n$ tori), it has a CW-complex structure made by a single 0-cell, $2n$ 1-cells 'wedged' to the 0-cell and a single 2-cell applied on the 1-cells. It is page 5 of Hatcher's Algebraic Topology.

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