Proof:

It clearly suffices to check the result on the generators of $S_n(X)$. So let $\sigma$ be an arbitrary singular $n$ simplex. Using equation (29.4),

$$(\partial _{n-1}\circ \partial _n)\sigma = \partial _{n-1}\Big( \... ...um_{i=0}^n\sum_{j=0}^{n-1}(-1)^{i+j} \sigma\circ(\Phi_i^n \circ \Phi_j^{n-1}).$$

To use lemma (29.1) we break the double sum in two pieces and write

$$(\partial _{n-1}\circ \partial _n)\sigma = \sum_{i\leq j}(-1)^{i+... ...hi_j^{n-1}) + \sum_{j < i}(-1)^{i+j} \sigma\circ(\Phi_i^n \circ \Phi_j^{n-1})$$

Using (29.2) in the second piece we get

$$(\partial _{n-1}\circ \partial _n)\sigma = \sum_{i\leq j}(-1)^{i+... ...^{n-1}) + \sum_{j < i}(-1)^{i+j} \sigma\circ(\Phi_j^n \circ \Phi_{i-1}^{n-1})$$

It may be noted that each of the two pieces is a sum of $n(n+1)/2$ terms (why?). Renaming $i-1$ as $k$ in the second sum gives

$$(\partial _{n-1}\circ \partial _n)\sigma = \sum_{i\leq j\leq n-1}... ...m_{j\leq k\leq n-1} (-1)^{k+j-1}\sigma\circ(\Phi_j^n \circ \Phi_{k}^{n-1}) = 0$$

as desired.

Now suppose that $X$ and $Y$ are two topological spaces and $f : X \longrightarrow Y$ is a continuous map then $f\circ \sigma$ is a singular $n-$simplex in $Y$ whenever $\sigma$ is a singular $n-$simplex in $X$.