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

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.

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.
• I have that if $f$ is constant then $f(e_0)=f(e_1)$ so $\partial(f)=f(e_1)-f(e_0)=0$ then $f\in Z_1(S_*(X))$ with which $[[f]]\in H_1(S_*(X))$, so according to this $H_1(S_*(X))=H_1(point)$? Why? – user402543 Jun 2 at 23:18