Many spaces we construct by drawing some shape and identifying sides. I want to know how this relates to the CW complex structure. I know that a CW complex structure is not unique but I would like to understand how to get a CW complex structure from such drawings.
For example, we can construct a torus like this
From the drawing on the left it seems that we have: four 0-cells, four 1-cells, and one 2-cell. This is clearly not correct as it gives the wrong Euler characteristic for the torus.
From the picture on the right its clear that we have: one 0-cell, two 1-cells, and one 2-cells. (This does gives the correct Euler characteristic.)
I know that the CW complex structure need not be unique, but its clear that my understanding for the CW complex on the left is wrong since it gives the wrong Euler characteristic.
My first idea was that maybe the CW complex on the left should have been:four 0-cells, 2 1-cells (since we are identifying opposite sides), and one 2-cell. I know that after we identify the sides $a$ and $b$, the four corners of the square are identified to one point, but I cannot see how we can add in the 1-cells without apriori having 4 0-cells.
The CW structure on the torus that is depicted on the right is built up inductively by dimension, as are all CW structures.
The $0$-skeleton is the un-named vertex where $\alpha$ and $\beta$ meet; let me call that vertex $V$.
The $1$-skeleton is $\alpha \cup \beta$, which can also be described as a quotient. There are two 1-cells to attach to the $0$ skeleton $V$. The 1-cell $\alpha$ is formed by taking some closed arc and identifying the endpoints of that arc with $V$. The 1-cell $\beta$ is formed by taking a disjoint closed arc, and identifying the endpoints of that arc also with $V$.
The $2$-skeleton is the whole torus, which can also be described as a quotient. There is one 2-cell to attach to the $1$ skeleton $\alpha \cup \beta$. The domain of that 2-cell is the square on the left side of the picture. The identification map is described by the labelling of the edges of that cell: each of the two vertical sides is identified with $\alpha$, and each of the two horizontal sides is identified with $\beta$.
Another way to express this is using a quotient map $q : S \to T$ where $S$ is the square and $T$ is the torus: the $0$-skeleton of $T$ is equal to the $q$-image of the vertex set of $S$; the $1$-skeleton of $T$ is equal to the $q$-image of the union of the sides of $S$; and the $2$-skeleton of $T$ is equal to the $q$-image of the whole of $S$, which is of course just the whole of $T$. This exact same description can be used to describe the skeleta of a CW complex structure on any surface that is constructed as the quotient of a polygon gluing diagram (e.g. the quotient of a "fundamental polygon").
Now, as to your questions, what the square depicts can also be described as a CW structure on the square itself: a $0$-skeleton with four $0$ cells; a $1$-skeleton with four $1$-cells; and a $2$-skeleton with one $2$-cell. The map from the square to the torus can be thought of as a CW map: each of the four $0$ cells of the square is mapped to the single $0$ cell $V$ of the torus; two of the four $1$-cells of the square are mapped to the $\alpha$ 1-cell of the torus; the other two of the four $1$ cells of the square are mpaped to the $\beta$ 1-cell of the torus; and the single $2$-cell of the square is mapped to the single $2$ cell of the torus.
In your question, you seem to be worried about what would happen if you attempted to start with the four $0$-cells of the square and then to attach the two $1$-cells of the torus to those four $0$-cells. That would indeed be a problem, but that's not what anyone would do. But one might, as described in the previous paragraph, simultaneously map all the cells of the square to the torus, and the problem you are worried about would not arise.