1

We know from Hatcher that ${H}_0(X,A) = 0$ iff $A$ meets every path component of $X$. A simpler case is that if $X$ is path-connected and $A$ is non-empty, then the first homology of the pair is zero.

I am not understanding where the following argument is wrong: $X=D^2$ is path-connected as contractible, $A=S^{1}$ is path-connected and meets the only path-component of $D^2$. Thus we should have $H_0(D^2, S^1)=0$ but we know that $D^2/S^1$ is homeomorphic to $S^2$, which has non-trivial $0-$th homology. Where do I go wrong?

EDIT: is it because the relative homology of a pair $(X,A)$ is not the homology of the quotient $X/A$?

thorwi
  • 57

1 Answers1

2

Under certain local niceness conditions (which are satisfied here), the relative homology of the pair $(X,A)$ is isomorphic to the homology of the quotient pair $(X/A,A/A)$ (an isomorphism is induced by the quotient map), which you can identify with the reduced homology of the space $X/A$, since $A/A$ is a point.

Thorgott
  • 17,265
  • by local niceness, do you mean that the pair $(X, A)$ is good? am I correct to say the following: if the pair is good, then we can use an LES of reduced homologies to determine the homology of the quotient $X/A$. On the other hand, if the pair is not good, then this LES does not hold. We must revert to the more general LES for any pair of spaces, where it is not true that $H_{i}(X,A)$ is the same as $H_{i}(X/A)$ in general. Thank you! – thorwi Jan 17 '22 at 12:32
  • Yes, the condition is that the pair is good. The only difference between the reduced and unreduced homology of the space is in degree $0$, where reduced homology is free on one generator less. This discrepancy is what your example observes. – Thorgott Jan 17 '22 at 15:33