What's the effect on the fundamental group of a path connected triangulable space X if you attach a disc?
So far I've figured out that a disc has trivial fundamental group. So the free product of their fundamental groups upon union will be isomorphic to the fundamental group of X, right? Then the fundamental group of the union is formed by adding the relations j = k where j, k are the inclusions of the intersection in each space. (van kampen)
From here i'm not sure how to proceed, I think I need to use that the space X is path connected or triangulable, somehow. Am i on the right track? This is from armstrong's topology by the way.
Using van Kampen's theorem, let $U=N(X)$ and let $V=N(D)$ where by $N(-)$ I mean 'take a small open neighbourhood'. You will see that $U\cap V$ is homotopy equivalent to a circle and so the fundamental group of $U\cup V$ is equal to $\pi_1(U)\ast_{\pi_1(U\cap V)}\pi_1(V)$ where we treat $\pi_1(U\cap V)$ as being the subgroup of $\pi_1(U)$ and $\pi_1(V)$ generated by the class of loops homotopic to the boundary circle $\partial D$.
This is isomorphic to $\pi_1(X)/\langle[\partial D]\rangle$ where here $[\partial D]$ denotes the 'class of maps homotopic to usual inclusion of the circle into the boundary of $D$'.