Proof:

Choose a point $x_0 \in X$ and $F:X\times [0, 1]\longrightarrow X$ be the homotopy $F(x, t) = (1-t)x + tx_0$. We shall define a group homomorphism $T:S_{n-1}(X)\longrightarrow S_{n}(X)$ satisfying a certain property (31.6) below. This is a special case of a chain homotopy that we shall encounter later in a more general context. Since $S_{n-1}(X)$ is a free abelian group generated by singular $(n-1)$ simplicies, it suffices to define $T$ these. For a singular $(n-1)$ simplex $\sigma:\Delta_{n-1}\longrightarrow X$, define the continuous map $T\sigma :\Delta_{n}\longrightarrow X$ in terms of the barycentric coordinates using the expression (31.3) namely

$$(T\sigma)(\lambda_1, \dots, \lambda_{n+1}) = F((\sigma\circ U)(\lambda_1, \dots, \lambda_{n+1}), \lambda_{n+1}).\eqno(31.4)$$

The continuity of $T\sigma$ is left as an exercise. Let us calculate the boundary of $T\sigma$ using equations (29.1) and (29.4). Recalling the notations used in lecture 29, one checks that $(T\sigma)\circ \Phi^n_n = \sigma.$

For $0 \leq j \leq n-1$, the $j-$th singular face is given by

$$(T\sigma)\circ \Phi^n_j(t_1, \dots, t_n) = F((\sigma\circ U)(t_1, \dots, t_{j-1}, 0, t_j,\dots, t_n), t_n). \eqno(31.5)$$

On the other hand when $0 \leq j \leq n-1$,

$$T(\sigma\circ\Phi^{n-1}_j)(t_1,\dots, t_n) = T\Big(\sigma\Big( \f... ...}}{1-t_n},0, \frac{t_j}{1-t_n},\dots,\frac{t_{n-1}}{1-t_n} \Big),\;t_n \Big),$$

which may be rewritten as $T((\sigma\circ U)(t_1, \dots, t_{i-1},0,t_i,\dots, t_{n-1}), t_n)$, in agreement with the right hand side of (31.5). From equation (29.4) it follows that for $n\geq 1$,

$$\partial _n(T\sigma) - T(\partial _n\sigma) = \sigma,\quad \sigma \in S_n(X), \eqno(31.6)$$

whereby we conclude that if $\sigma \in Z_n(X)$ then $\sigma = \partial _n(T\sigma) \in B_n(X)$. That is $Z_n(X) = B_n(X)$.