I came across the fact the torus i.e. $S^1 \times S^1$ has the structure of a CW-complex. To see that we need to start with the $0$-skeleton $X^0$ consisting of a single point $*.$ Then we attach two $1$-cells to $X^0$ via the obvious attaching map $* \bigsqcup *$ from $S^0 \bigsqcup S^0$ to $X^0.$ With this attachment map the pushout we obtain is the $1$-skeleton $X^1$ which is the wedge of two circles $S^1 \vee S^1.$ At last we need to to attach a single $2$-cell to $X^1$ to get a torus. But here it is not known to me that what would be the attaching map from $S^1$ to $S^1 \vee S^1.$ Do anybody have any idea about that? A small hint will be highly appreciated at this stage.
Thanks for your time.
Best Answer
Separate $S^1$ in four arcs call them $p_1p_2,p_2p_3,p_3p_4,p_4p_1$ and direction is clockwise and identify $D^2$ as $I^2$ , mark this square square as ABCD in an anticlockwise direction starting from left. Let us also call the two circles of $S^1\vee S^1$ by $a,b$ with appropriate direction (think them as a directed path). Now you map the arc $\overrightarrow{p_1p_2}$ to side $\overrightarrow{AB}$ of the square , the arc $\overrightarrow{p_2p_3}$ to side $\overrightarrow{BC}$ of the square and so on. Now, we give the identification map as follows, map the arc $\overrightarrow{p_1p_2}$ to the directed loop $a$ then the arc $\overrightarrow{p_2p_3}$ to the directed loop $b$ , then the arc $\overrightarrow{p_3p_4}$ to the directed loop $a^{-1}$ (i.e. now traverse in the opposite direction) and finally the arc $\overrightarrow{p_4p_1}$ to the directed loop $b^{-1}$ and this gives you the torus as the initial object of the pushout diagram