Let $H$ be a homology theory satisfying Eilenberg-Steenrod axioms and $X$ an arbitrary topological space. We can write $X$ as a disjoint union of its points $$X= \coprod_{x \in X}{\{x\}}$$ Now the additivity axiom implies that $H_n(X)$ is isomorphic to $\bigoplus_{x \in X } H_n(\{x\})$. By the dimension axiom, we have that $H_n(X)=0$ for $n\neq0$.

That generally $H_n(X)=0$ for $n\neq0$ holds is of course wrong (just look at homology groups of the 2-sphere). My question is now:

Where is the wrong argument in the first paragraph?

  • 2
    $\begingroup$ As a set, X is a disjoint union of its points, but not necessarily so as a topological space. $\endgroup$ – fixedp Feb 9 '16 at 19:54

Eilenberg-Steenrod axioms says that, if $X= \coprod_{\alpha} X_\alpha$ the disjoint union of family of topological spaces $X_\alpha$ then $H_n(X)= \bigoplus H_n(X_\alpha)$ , so here each $X_\alpha$ is open in $X$. But the family you have defined, in that case each singleton $x\in X$ may not be an open set in $X$, and only possibility is in case of discrete space.

| cite | improve this answer | |
  • $\begingroup$ Thanks for the quick answer. Of course, it is meant the disjoint union of topological spaces not only of sets. $\endgroup$ – bjn Feb 9 '16 at 20:10

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.