0
$\begingroup$

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

enter image description here

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.