$H_0 ( X )$ is free abelian on the path components of $X$ .

Question

I do not quite understand this statement. Does this mean if there is one path-components, then $H_0 ( X )$ is generated by one generator, or cyclic. And if there is two path-components, then $H_0 ( X )$ is generated by two generator, but of the form $a^i b^j$?

And why this is true?

As to why this is true, see proposition 2.7 in Hatcher's Algebraic Topology (a previous post of yours suggests that you're reading it). — Ayman Hourieh, Nov 03 '13 at 16:08

score 1 · Accepted Answer · answered Nov 03 '13 at 16:02

It is customary to write homology groups additively rather than multiplicatively. So if you mean $a$ and $b$ to be the generators, then an arbirary element will be a linear combination of $a$ and $b$ with arbitrary coefficients (presumably you mean integer coefficients). Other than that your guess is correct: the number of generators is the number of path-connected components.

$H_0 ( X )$ is free abelian on the path components of $X$ .

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