(June ’05) Find a cell structure for the quotient space obtained by identifying two distinct points \(a,b\) in a \(2\)-torus to a third point \(c\) in a \(2\)-sphere, and compute a presentation for the fundamental group of this space.
Solution
Let \(T=T^2\). Subdivide \(T\) so that \(a\) and \(b\) are vertices and choose an edge path from \(a\) to \(b\). After identifying \(a\sim b\), that path becomes one additional loop. Collapsing a maximal tree in the resulting \(1\)-skeleton gives a CW model with one \(0\)-cell, three \(1\)-cells \(x,y,t\), and one torus \(2\)-cell attached by the commutator word \[ [x,y]=xyx^{-1}y^{-1}. \]
The sphere meets the torus quotient only at the common point \(c\). Give \(S^2\) its CW structure with one \(0\)-cell and one \(2\)-cell attached by the constant map. Thus the whole quotient has a CW model with one \(0\)-cell; \(1\)-cells \(x,y,t\); a \(2\)-cell attached by \([x,y]\); and a further \(2\)-cell attached trivially (the sphere).
The trivial sphere attaching map adds no relation to \(\pi_1\). Therefore \[ \pi_1(X)\cong\langle x,y,t\mid [x,y]=1\rangle \cong \mathbb Z^2*\mathbb Z. \]