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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 http://books.google.com/books?id=zuWttfJmexwC&pg=PA32&lpg=PA32&dq=%22construction+of+the+torus%22+cells&source=bl&ots=B7xdOKizBQ&sig=yE_shMjNNTBzb3SnmYbTD766cF0&hl=en&ei=ZpUFTeCgC4GClAeDmM3HCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBcQ6AEwAA#v=onepage&q=%22construction%20of%20the%20torus%22%20cells&f=false help?2010-12-13
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    http://www.youtube.com/watch?v=0H5_h-RB0T82010-12-13
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    Arturo - wish it did, but no I'm still having trouble seeing it. (Btw I found the whole article at http://math.rice.edu/~forman/morseintro.ps and am still working through it. The dynamical systems bits seem extraneous to what I'm trying to understand, but interesting nonetheless.) I'm working from the understanding of a cell as anything (in 2 or 3 dimensions, I suppose) resulting from the continuous transformation of a (2 dimensional) disk.2010-12-13
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    Qwirk - wow, cool video. But I'm working from the (perhaps not universal?) understanding that a cell should not be glued to itself in constructing a new surface. And it seems as if this video demonstrates a torus constructed from a cell glued to itself, no?2010-12-13
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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).2010-12-13
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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.2010-12-13
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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.2010-12-13
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    @EricO.Korman From the video: To clarify, what is the 2-cell? My understanding is that an n-cell is homeomorphic to an n-ball. (Is that correct? Is this the precise definition?) So is the 2-cell the original rectangle (which is homeomorphic to a 2-ball?) and by bending the rectangle as done in the video, its boundary is being "attached" along two blue arcs (the 1 cells) and then joined at a single point (the 0 cell)?2016-02-22

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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.