$f$ is a cycle since it is a loop, and since $H_1(point) = 0$, $f$ must then be a boundary.

Question

I am reading Hatcher's book (algebraic topology, p.166) and I can not understand what he says in the book:

I know that $H_1(point)=0$, but I do not know why this implies that "$f$ must then be a boundary". Could someone please help me? Thank you.

Answer 1

This is literally the definition of $H_1$. By definition, $H_n(X)$ is the quotient of the group of $n$-cycles by the subgroup of boundaries. So since $H_1(point)$ is trivial, that means every $1$-cycle in $point$ is a boundary. In particular, $f$ (which as a constant map can be considered as taking values in $point$) is the boundary of some $2$-chain in $point$ which then can also be considered as a $2$-chain in the original space.