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

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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.

  • $\begingroup$ 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? $\endgroup$ – user402543 Jun 2 at 23:18

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