Exercise 1.2.14

Proof. We first draw the $2$-cell with the given identifications:

where edges with the same label are identified, the vertices $\text{∙}$ are identified, as are the vertices $\text{⋆}$, and opposite faces are identified. These edge identifications follow just by considering opposite faces and using the one-quarter twist identification, and by matching heads/tails of edges to the vertices. Since this cube is then filled with a 3-cell, we see that $X$ is a cell complex with $0$-cells, four $1$-cells, three $2$-cells, and one $3$-cell.

Now we calculate the fundamental group. Recall that attaching $n$-cells for $n>2$ in a CW decomposition has no effect on the fundamental group, so we only have to consider the fundamental group for the $2$-skeleton drawn above.

To calculate this fundamental group, we first deformation retract $d$ to a point. Thus, our 2-cells attach according to the schema $a{c}^{-1}{b}^{-1}$, $c^{-1}b{a}^{-1}$, and $b{c}^{-1}a$, and we have the presentation

Letting $a=i$, $b=j$, $c=k$ this is equivalent to the presentation

We can multiply $i=\mathit{jk}$ by $i$ and $\mathit{ij}=k$ by $k$. Substituting $i=\mathit{jk}$ into $j=\mathit{ki}$ to get $j=\mathit{kjk}$, multiplying by $j$ to get $j^2=\mathit{jkjk}$, and then substituting $i=\mathit{jk}$ again to get $j^2=\mathit{ijk}$, we finally have the presentation

which is the usual presentation for the quaternion group. □