What is the fundamental group of $X=S^2\cup\text{a disc parallel to the plane of equator}$? How I imagine it:
I think a cell structure of $X$ consists of 1 0-cell, 1 1-cell (the equator)-call it $a$, 3 2-cells (north-south hemispheres and the disc). But I'm not sure about the 2-cells,  there may be only one 2-cell with its boundary divided into 3 pieces and attached to $a$ according to the word $aaa^{-1}$? Or are these the same things? I think $X$ has the trivial fundamental group. Can someone explain these in detail?
 A: The answer Mariano gave is the way I would tackle the problem because it's easy enough to see that the space is homotopy equivalent to the wedge of two spheres and, as spheres are locally contractible, Van Kampen's theorem tells us that the fundamental group of the space is isomorphic to the free product of the fundamental groups of each sphere - which are trivial.
That being said, proving this by cellular decomposition is just as easy. Let $X$ be the space in the question. We can view $X$ as a CW-complex with one $0$-cell a point on the equator, one $1$-cell the equator, and three $2$-cells which are the two hemispheres and disc, all attached to the equator wrapping around one on their boundary.
The $1$ cells are the generators of the fundamental group and so we have a single generator given by wrapping around the equator. Call this generator $a$ with representative $\phi\colon S^1\rightarrow X$. Consider any of the three $2$-cells. They are all attached to the one cell via the map $\phi$ on their boundary and so in particular, as a loop on the boundary of the cell can be contracted to a point as the $2$-cell is contractible, we deduce that $a$ is also a relator in the fundamental group and so $\pi_1(X)=\langle a\mid a\rangle\cong 1$.
