Proof:

Let $\iota_p$ be the element of $A_p(\Delta_p)$ given by the identity map from $\Delta_p$ to itself. Since an arbitrary $\sigma\in S_p(X)$ can be written as $\sigma = \sigma_{\sharp}\iota_p$, condition (i) forces

$${\cal B}\sigma = ({\cal B}\circ\sigma_{\sharp}) \iota_p = \sigma_{\sharp}({\cal B}\iota_p) = \sigma_{\sharp}B\iota_p \eqno(34.10)$$

since ${\cal B}\iota_p = B\iota_p$ by (ii). Thus the conditions (i) and (ii) determine ${\cal B}$ uniquely on the generators of the free abelian group $S_p(X)$. The same argument applies to ${\cal J}$.

We use (34.10) and (34.5) to show that $B$ and ${\cal B}$ agree on any affine simplex $\sigma \in A_p(\Delta_n)$. Denoting by ${\bf g}$ the barycenter of $\iota_p$ and by ${\bf b}$ the barycenter of $\sigma$,

$${\cal B}\sigma = \sigma_{\sharp}B\iota_p = \sigma_{\sharp}(K_{\bf g}(\partial _p\iota_p)).$$

Using exercise 2, this may be rewritten as

$${\cal B}\sigma = K_{\sigma({\bf g})}(\sigma_{\sharp}(\partial _p\... ...}(\partial _p\sigma_{\sharp}\iota_p) = K_{\bf b}(\partial _p\sigma) = B\sigma.$$

The verification for ${\cal J}$ is similar. We now run through the proof that ${\cal B}$ is a chain map, which is now automatic. For an arbitrary $\sigma\in S_p(X)$, $\partial {\cal B}\sigma = \partial (\sigma_{\sharp}B\iota_p) = \sigma_{\sharp}(\partial B\iota_p).$ Since $\partial \iota_p$ is an affine chain and $B$ is a chain map on the subcomplex of affine chains we get $\partial B\iota_p = B\partial \iota_p.$ Applying $\sigma_{\sharp}$ to this gives $\partial {\cal B}\sigma = \sigma_{\sharp}(B\partial \iota_p)$. Working from the other end using the fact that $\sigma_{\sharp}$ is a chain map and ${\cal B}$ satisfies (i), we get

$${\cal B}\partial \sigma = {\cal B}\partial (\sigma_{\sharp}\iota_... ...igma_{\sharp}{\cal B}(\partial \iota_p) = \sigma_{\sharp}B(\partial \iota_p).$$

Finally we show that ${\cal J}$ is a chain homotopy between ${\cal B}$ and the identity operator. For $\sigma\in S_p(X)$,

$${\cal J}\partial \sigma = {\cal J}\partial \sigma_{\sharp}\iota_p... ..._p = \sigma_{\sharp}{\cal J}\partial \iota_p = \sigma_{\sharp}J\partial \iota_p$$

$$\partial {\cal J}\sigma = \partial {\cal J}\sigma_{\sharp}\iota_p... ...p = \sigma_{\sharp}\partial {\cal J}\iota_p = \sigma_{\sharp}\partial J\iota_p.$$

We have used (i) and (ii) and the fact that $\sigma_{\sharp}$ is a chain map. Subtracting and using (34.9) we get the desired result. $\square$