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My question is about tom Dieck's proof of Theorem 9.2.1 on page 227, which states that if $X$ is path connected, then the induced map $$h:\pi_1(X,x_0)^{ab}\to H_1(X)$$ is an isomorphism. Specifically, showing that the above map is injective. Here is the passage:

We construct a homomorphism in the other direction. For $x\in X$ we choose a path $u(x)$ from $x_0$ to $x$. We assign to a 1-simplex $\sigma:\Delta^1\to X$ from $\sigma_0=(1,0)$ to $\sigma_1 = (0,1)$ the class of the loop $u(\sigma_0)\ast\sigma\ast u(\sigma_1)^{-1}$. We extend this assignment linearly to a homomorphism $l':C_1(X)\to\pi_1(X,x_0)^{ab}$. Let $\tau:\Delta^2\to X$ be a 2-simplex with faces $\tau_j=\tau d_j$. Since $\Delta^2$ is contractible, $\tau_2\ast\tau_0\simeq\tau_1$. This implies $$l'([\tau_2])+l'([\tau_0]) = l'([\tau_2+\tau_0]) = l'([\tau_2\ast\tau_0]) = l'([\tau_1]).$$ Hence $l'$ factors over $C_1(X)/B_1(X)$ and induces $l:H_1(X)\to\pi_1(X,x_0)^{ab}$. By construction, $lh=id$.

I'm ok with all of it except the string of equalities. By definition, $l'$ is a map from the 1-cycles, but it seems to be taking elements of $H_1(X)$ as inputs. Furthermore, if I'm not mistaken, the $\tau_j$ aren't even 1-cycles? I've tried to reconcile this by looking at the image of $\tau_0-\tau_1+\tau_2\in C_1(X)$, but we still need to work in homology classes to utilize $[\tau_0+\tau_2]=[\tau_0\ast\tau_2]$. My question is, why does $l'$ take homology classes as inputs?

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    $\begingroup$ I disagree; from the definition, $l'$ eats all chains, not just cycles. He shows $l'$ is trivial on boundaries, and so the restriction of $l'$ from all chains to cycles induces a map out of homology, as claimed. $\endgroup$
    – hunter
    Jun 25, 2016 at 20:45
  • $\begingroup$ @hunter That makes perfect sense. But why is $l'$ taking homology classes as inputs when it's defined on chains? It's my understanding that $[\tau_0-\tau_1+\tau_2]$ is an element of $C_1(X)/B_1(X)$ $\endgroup$
    – user171308
    Jun 25, 2016 at 20:59
  • $\begingroup$ There appear to be a lot of typos in the passage you have quoted. You are of course correct that $l'$ is defined on chains, rather than on homology classes. The formula $u(x)\ast\sigma\ast u(x)^{-1}$ also doesn't make sense: what is $x$? Presumably the first $x$ should be $\sigma_0$ and the second $x$ should be $\sigma_1$ $\endgroup$ Jun 25, 2016 at 22:03

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This is just a notational error. It should say $$l'(\tau_0)+l'(\tau_2) = l'(\tau_0+\tau_2) = l'(\tau_0\ast\tau_2) = l'(\tau_1).$$ To show that $l'(\tau_0+\tau_2)=l'(\tau_0\ast\tau_2)$, you just directly use the definition of $l'$: $$l'(\tau_0+\tau_2)=[u(x)\ast\tau_0\ast u(y)^{-1}\ast u(y)\ast\tau_2\ast u(z)^{-1}]=[u(x)\ast\tau_0\ast\tau_2\ast u(z)^{-1}]=l'(\tau_0\ast\tau_2),$$ where $x$ and $y$ are the endpoints of $\tau_0$ and $y$ and $z$ are the endpoints of $\tau_2$. Here the brackets indicate taking equivalence classes of loops to get elements of $\pi_1(X,x_0)^{ab}$.

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