Proof:

If $p = 0$ and $\sigma$ is a zero chain then $B\partial \sigma = 0$ whereas $\partial B\sigma = \partial \sigma = 0.$ To handle the case $p > 0$ we assume inductively that for any $k-$chain $\sigma$ with $k\leq p-1$, the equation $\partial B\sigma = B\partial \sigma$ holds. To prove it for $p$ chains, let $\sigma$ be an arbitrary affine $p-$simplex. Equations (34.5) and (34.3) combine to give

$$\partial B\sigma = \partial K_{\bf b}(B\partial \sigma) = B\parti... ...) = B\partial \sigma - K_{\bf b}(B\partial \partial \sigma) = B\partial \sigma.$$

Note that induction hypothesis justifies $\partial B\partial \sigma = B\partial \partial \sigma$. We have now shown that for every $p$ chain $\sigma$,

$$B\partial \sigma = \partial B\sigma. \eqno(34.6)$$

We now construct a chain homotopy $J:A_p(\Delta_n)\longrightarrow A_{p+1}(\Delta_n)$ between $B$ and the identity map. Equation (34.3) suggests a formula of the type

$$J\sigma = K_{\bf b}f(\sigma),$$

where ${\bf b}$ is the barycenter of $\sigma$ and $f:A_p(\Delta_n)\longrightarrow A_p(\Delta_n)$ is to be determined. The condition that $J$ is a chain homotopy between $B$ and the identity now forces

$$f(\sigma) - K_{\bf b}\partial f(\sigma) = B\sigma - \sigma - J(\partial \sigma). \eqno(34.7)$$

Clearly $f(\sigma) = 0$ for a zero simplex $\sigma$. If we assume inductively that $J:A_k(\Delta_n)\longrightarrow A_{k+1}(\Delta_n)$ has already been defined for $k\leq p-1$, the right hand side of (34.7) is then a known function. Let us refer to the term $K_{\bf b}\partial f(\sigma)$ in equation (34.7) as junk. Exercise 3 invites the reader to check that retaining the junk term is unnecessary. We set it equal to zero and define formally for a $p-$simplex $\sigma$,

$$J\sigma = \left\{\begin{array}{lll} 0 & & \mbox{ if } p = 0, \\ ... ...sigma - J(\partial \sigma)) & & \mbox{ if } p \geq 1. \\ \end{array} \right.$$

Let us now verify that this formula does the job. The case $p = 0$ is trivial and let us assume

$$\partial J\sigma + J(\partial \sigma) = B\sigma - \sigma$$

for any $k$ chain such that $k\leq p-1$. Using the formula of $J$ we see that

$$\partial J\sigma = B\sigma - \sigma - J(\partial \sigma) - K_{\bf... ...\partial B\sigma - \partial \sigma - \partial J(\partial \sigma)). \eqno(34.8)$$

By induction hypothesis $\partial J(\partial \sigma) = -J(\partial \partial \sigma) + B(\partial \sigma) - \partial \sigma$. Inserting this in (34.8) we get the desired result

$$\partial J\sigma = B\sigma - \sigma - J(\partial \sigma). \eqno(34.9)$$

We shall now transfer the barycentric subdivision operator and the chain homotopy $J$ to a chain map ${\cal B}:S_p(X)\longrightarrow S_p(X)$ and a chain homotopy ${\cal J}:S_p(X)\longrightarrow S_{p+1}(X)$. This will be unique subject to naturality.