I need to compute the fundamental group (and other homotopy groups) of the circle with the chord (or diameter, it doesn't matter).
If I understand all correctly, then this topological space is homotopy equivalent to the "eight" (two circles with one common point).
Then, the fundamental group is isomorphic to free group with two generators and similarly we can compute other homotopy groups.
Does these two spaces are homotopy equivalent?
Best Answer
Yes, the two spaces are homotopy equivalent. The point is that the line segment is contractible in your space.
More formally, let $X$ be this space and give it a cell structure with $2$ $0$-cells and $3$ $1$-cells, where each $1$-cell is attached to the $0$-skeleton by putting one endpoint at each of the $0$-cells. Then the diameter $A$ is a subcomplex of $X$ ($A$ consists of two $0$-cells and a single $1$-cell), and $A$ is contractible. Then it's well-known that the quotient $X \to X/A$ is a homotopy equivalence (e.g., Hatcher Proposition 0.17), but $X/A \cong S^1 \vee S^1$.