Proof:

The construction proceeds by induction on the total degree $p+q$ on pairs $(\sigma, \tau)$. It has already been carried out for the case when one of $\sigma$ or $\tau$ is a zero simplex and in particular when the total degree $p+q$ is zero. Further, and for this case, conditions (ii) and (iii) hold trivially. Assume that the cross product

$$S_p(X)\times S_q(Y)\longrightarrow S_{p+q}(X\times Y) \eqno(33.3)$$

has been defined for all pairs $(p, q)$ such that $p+q < k$ satisfying (ii) and (iii). Now if $\sigma\in S_p(X)$ and $\tau\in S_q(Y)$ are such that $p+q = k$ then the right hand side of the formula in (iii) already makes sense and in particular this is so with the pair $\iota_p$ and $\iota_q$. Thus we need a singular $p+q$ chain $z$ such that

$$\partial z = \partial \iota_p \times \iota_q + (-1)^p(\iota_p\times \partial \iota_q). \eqno(33.4)$$

Applying Leibnitz rule again to the right hand side one checks that it is a cycle. Since $\Delta_p\times\Delta_q$ is convex this cycle is also a boundary and so (33.4) has a (non-unique) solution $z \in S_{p+q}(\Delta_p\times\Delta_q)$. Once this choice is made the construction proceeds further as follows. Each $\sigma\in S_p(X)$ can be realized as $\sigma_{\sharp}(\iota_p)$ where $\sigma:\Delta_p\longrightarrow X$ and likewise for a singular $q$ simplex $\tau$ in $Y$. But now equation (33.1) forces upon us the definition

$$\sigma\times \tau = \sigma_{\sharp}(\iota_p)\times \tau_{\sharp}(... ...\tau)_{\sharp}(\iota_p\times \iota_q) = (\sigma\times \tau)\circ z,\eqno(33.5)$$

where the $\sigma\times \tau$ appearing on the extreme left of (33.5) is the object we are defining whereas the $\sigma$ and $\tau$ appearing in the middle and on the extreme right of (33.5) denote the functions $\sigma:\Delta_p\longrightarrow X$ and $\tau:\Delta_q\longrightarrow Y$. The easy verification of (33.1) is left for the reader. Proof of (33.2) runs as follows:

$$\partial (\sigma\times\tau) = \partial \big( (\sigma\times\tau)_{\sharp}(\iota_p\times \iota_q)\big)$$

$$= (\sigma\times\tau)_{\sharp}\partial (\iota_p\times \iota_q)$$

$$= (\sigma\times\tau)_{\sharp}\partial z$$

$$= (\sigma\times\tau)_{\sharp}\Big( \partial \iota_p \times \iota_q + (-1)^p(\iota_p\times \partial \iota_q)\Big)$$

Applying (33.1), which holds by induction hypothesis, and using the pair of equations $\sigma_{\sharp}\partial = \partial \sigma_{\sharp}$, $\tau_{\sharp}\partial = \partial \tau_{\sharp}$ we continue with our calculation:

$$\partial (\sigma\times\tau) = \sigma_{\sharp}(\partial \iota_p)\times \tau_{\sharp}(\iota_q) + (-1)^p(\sigma_{\sharp}(\iota_p)\times\tau_{\sharp}(\partial \iota_q))$$

$$= \partial\sigma\times\tau + (-1)^p(\sigma\times \partial \tau).$$

Having defined $\sigma\times \tau$ for singular simplices $\sigma$ and $\tau$, we can extend it as a bilinear map $S_p(X)\times S_q(Y)\longrightarrow S_{p+q}(X\times Y)$ since $S_p(X)$ and $S_q(Y)$ are free abelian groups.