# Calculating H_0 directly from Eilenberg-Steenrod axioms

It's well-known that every homology theory satisfying Eilenberg-Steenrod axioms is isomorphic to singular homology. I tried to perform some homology calculations directly from axioms but couldn't do even a simple task: I can't prove that $H_0$ of a (linearly) connected space is isomorphic to $H_0$ of a point. It's not hard to prove (by considering morphism of exact sequences of pairs $(X, *) \to (*, *)$) that the map $H_0(*) \to H_0(X)$ induced by inclusion of a point is a monomorphism, and $X$ being linearly connected implies that any such inclusion induces the same map (they are homotopic). I see no way to use this fact to prove surjectivity.

Calculation of $H_0$ looks by an order of magnitude simpler than proving the uniqueness of homology theory, so I hope there is a quick and elementary argument for it. Could someone help me?

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I don't know a better way than: 1. prove from the axioms that a CW-complex with $1$ vertex, $Y$, satisfies $H_0(Y)= \mathbb Z$. 2. Show that if $X$ is path connected there is a CW-complex with $1$ vertex $Y$, and a weak equivalence $Y\to X$. –  Justin Young Jan 25 '13 at 13:16
I think your argument only works if $X$ has the homotopy type of a CW-complex. Because if it doesn't, I think that the weak homotopy equivalence $Y \to X$ does not necessarily induce an iso in homology. –  redfiloux Jan 26 '13 at 7:24
Similar question on MO : mathoverflow.net/questions/41299/… –  redfiloux Jan 27 '13 at 1:03
Yes, I was using my own version of the ES axioms, which apparently is not the standard one. You can at least show that for spaces of the homotopy type of CW complexes, your result holds, and there may be some kind of hint of a counterexample in the MO answer. –  Justin Young Jan 28 '13 at 11:19