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I'm trying to show directly that $$ H_0(X, x_0) \cong \widetilde{H}_0(X) $$ where $x_0$ is a point in the topological space $X$ and $\widetilde{H}_0$ denotes the zeroth reduced (singular) homology group.

What I have so far gives me a wrong answer, I'd be grateful if anyone can point out the mistake and the right direction to go.

Attempted proof: As usual we have the chain complex $$ \cdots \longrightarrow \frac{C_1(X)}{C_1(x_0)} \stackrel{d_1}{\longrightarrow} \frac{C_0(X)}{C_0(x_0)} \longrightarrow 0$$ and by definition $$ H_0(X,x_0) = \frac{C_0(x)/C_0(x_0)}{\text{im}(d_1)}. $$

Now $$d_1\left(\frac{C_1(X)}{C_1(x_0)}\right) = \frac{d_1(C_1(X))}{C_0(x_0)}, \tag{$\ast$}$$ so $$ H_0(X,x_0) \cong \frac{C_0(X)}{d_1(C_1(X))} $$ by the third isomorphism theorem.

$C_0(X)$ is the free abelian group generated by the points of $X$, and $d_1$ sends the generators $$ \sigma : [0,1] \rightarrow X$$ of $C_1(X)$ to $(\sigma(1) - \sigma(0))$, so $$ d_1(C_1(X)) = \mathbb{Z}\{y-x\ |\ x,y \ \text{in the same path component of}\ X\},$$ and after modding out $C_1(X)$ by this we get that $H_0(X,x_0)$ is a direct sum of $\mathbb{Z}$'s, one for each path component of $X$. ($\ast\!\ast$)

But we know that $\widetilde{H}_0(X)$ is a direct sum of $n$ copies of $\mathbb{Z}$, where $n$ is one less than the number of path components of $X$. (For $X$ having a finite number of path components.)

There are two places in this argument I'm a bit suspicious of, namely the bits marked ($\ast$) and ($\ast\!\ast$), but I can't quite say what the problem might be. Help please?

Note that the same question is answered using chain homotopies here, but I'm looking for a more direct proof.

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Josh
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1 Answers1

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There is a problem with $(*)$: remark that $C_0(x_0)$ is not even a subgroup of $d_1(C_1(X))$, so it makes no sense to write the quotient. It's not a subgroup because $C_0(x_0)$ contains elements with nonzero "weight" (i.e. things which aren't in the kernel of the augmentation map). However, the image of $C_1$ lies in the kernel of the augmentation map. I'll let you think about what the image really is.

As for $(**)$, it's a little vague, so I hope that once you take care of $(*)$ you'll be able to clear it up as well.

Here's a general tip: instead of trying to prove that two gadgets are isomorphic by computing them side by side, focus on constructing a map from one gadget to the other, and then prove that your map is an isomorphism.

Bruno Joyal
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