# Why does a CW-complex consist of *open* cells?

Wikipedia states

A CW complex is a Hausdorff space X together with a partition of X into open cells (of perhaps varying dimension) that satisfies two additional properties:

1. For each n-dimensional open cell C in the partition of X, there exists a continuous map f from the n-dimensional closed ball to X such that

• The restriction of f to the interior of the closed ball is a homeomorphism onto the cell C, and

• the image of the boundary of the closed ball is contained in the union of a finite number of elements of the partition, each having cell dimension less than n.
2. A subset of X is closed if and only if it meets the closure of each cell in a closed set.

I am confused by this definition, because I don't see how for example a torus is a CW complex (I've been told it is), given that the cells have to be open.

Here's my problem: I'd consider a torus to consist of one open 2-cell (the image of an open square/disk onto the main shape of the torus), and two 1-cells, namely the left and right side of the square glued together and the top and bottom sides glued together.

The problem with this is, because the sides have to be open cells, we havent included the corners of the square. If we would include that, it wouldn't be an open cell anymore. So in this construction, either one or two points of the torus are not included in the partitioning into cells.

What mistake am I making?

• The corners are $0$-cells... – YuiTo Cheng Apr 27 at 10:48
• @YuiToCheng, oohhh – user56834 Apr 27 at 10:59

## 1 Answer

each open cell is open in its own dimension. So an open 0-cell is a point. let's call our point $$v$$. If you glue two closed 1-balls, call them $$a$$ and $$b$$ (closed interval) onto $$v$$ along their boundary, we get a wedge of two circles, where the basepoint is $$v$$ (and the endpoints of the intervals are also $$v$$). Thus the open 1-cell $$a$$ is homeomorphic to an open petal of the rose (so take away $$v$$).

Finally, glue in a closed $$2$$-ball (a disk) so that the boundary of the $$2$$ disk traces out $$abAB$$, where implicitly I have chosen an orientation of $$a$$ and $$b$$ and indicated $$A$$ as meaning tracing out $$A$$ in the reverse orientation. From my description it may not be obvious that the CW-complex you get is a torus, but I claim that it is.

The schematic is this. Draw a square, label the corners (0-cells) $$v$$, and then label and orient the edges $$a$$ and $$b$$ so that starting at the initial vertex of an $$a$$ edge and walking forward around the square we read $$a$$, then $$b$$, then $$a$$ with the opposite orientation and $$b$$ with the opposite orientation. Now glue oriented edges together. You should get a torus!