 Solution

The Seifert-van Kampen theorem states that if a space is covered by path-connected open sets $U,V$ with path-connected intersection containing a basepoint, then its fundamental group is the free product ${\pi }_1(U)\ast {\pi }_1(V)$ modulo the normal subgroup generated by identifying the two images of each loop in $U\cap V$. More generally it is the corresponding pushout of groups.

A wedge of two circles has fundamental group $F(a,b)$, the free group on their two loops: apply the theorem to thickened circle neighborhoods whose intersection is contractible and use ${\pi }_1(S^1)=\mathbb{Z}$. Represent the torus by a square with opposite edges identified. Remove an interior disk; the remainder deformation retracts onto this wedge. Replacing the disk attaches a two-cell whose boundary follows $ab{a}^{-1}{b}^{-1}$. A disk neighborhood and the punctured torus have annular intersection, and van Kampen kills precisely that boundary loop. HenceA torus with a point removed retracts to a torus with a small open disk removed, and then to the same wedge. ThusFor the connected sum, remove a disk from each torus and glue their boundary circles with reversed orientations. The two pieces have free groups $F(a,b)$ and $F(c,d)$ and their overlap is an annulus. Its boundary generators correspond to $[a,b]$ and $[c,d{]}^{-1}$, where $[a,b]=ab{a}^{-1}{b}^{-1}$. ThereforeThe boundary relation is essential: this group is neither the free group on four generators nor the abelian group ${\mathbb{Z}}^4$.