Why natuality carries the special case to the general case in Eilenberg–Zilber theorem? This question is from the paper Eilenberg–Zilber via acyclic models, and products in homology and cohomology by Chris Kottke at the end of Proposition 1.6.
(Eilenberg–Zilber)
show that there exist a natural map $D \colon (S_*(X)\otimes S_*(Y))_* \to(S_*(X)\otimes S_*(Y))_{*+1}$ such that $\mathrm{Id} - \theta x = \partial D + D \partial$ where $\mathrm{Id}$ be identity map and
$$
  x \colon (S_*(X) \otimes S_*(Y))_* \to S_*(X \times Y)
  \quad\text{by}\quad
  x(\alpha \otimes \beta) = (\alpha \times \beta)_{\#} x(i_p \otimes i_q)
$$
and
$$
  \theta \colon S_*(X \times Y) \to (S_*(X) \otimes S_*(Y))_*
  \quad\text{by}\quad
  \theta(\sigma) = (\pi_X \sigma)_{\#} \otimes (\pi_Y \sigma)_{\#}\theta(d_n)
$$
where $d_n \colon \Delta^n \to \Delta^n \times \Delta^n$ is an $n$-simplex.
I have understand the “acyclic model” for the case for $X = \Delta^p$, $Y = \Delta^q$ and I know that when $D$ is restriction to $i_p \otimes i_q$ it will satisfy this equality $\mathrm{Id} - \theta x = \partial D + D \partial$, but I don’t understand is the final word in the end of this propositon. It says “since $D$, $\mathrm{Id}$, $\theta x$, and $\partial$ are also natural in $X$ and $Y$, so we can extend the equality, $\mathrm{Id} - \theta x = \partial D + D \partial$, to general space $X, Y$”. How can this be true? I know the natural condition will hold, but how can we use natural property to general space?
If I am not misunderstanding the question, I will need this proof:
(Note that the upper diagram in the lemma $X=\Delta^p$,$Y=\Delta^q$ (I forget to remind, and the below X,Y be arbitrary space.)

Can someone give me a hint to prove that?
 A: Fortunatly, I got the answer. The key is that all be commutative diagram. Since I took two blank sheets of paper to complete this proof, if the following sketch of an answer does not satisfy you, please tell me, and I will post my complete answer along with these papers.
Let $\sigma\in S_*(X\times Y);d_n$ be diagonal map
$$
\begin{split}
D\partial\sigma&=D\partial[\pi_X\sigma\times\pi_Y\sigma]_{\#}(d_n)\\
&=D[\pi_X\sigma\times\pi_Y\sigma]_{\#}\partial(d_n)\\
&=[\pi_X\sigma\times\pi_Y\sigma]_{\#}D\partial (d_n)\\
&=[\pi_X\sigma\times\pi_Y\sigma]_{\#}[(Id-\theta x)(d_n)-\partial  D(d_n)]\\
&=[\pi_X\sigma\times\pi_Y\sigma]_{\#}(Id-\theta x)(d_n)-[\pi_X\sigma\times\pi_Y\sigma]_{\#}\partial D(d_n)\\
&=(Id-\theta x)[\pi_X\sigma\times\pi_Y\sigma]_{\#}(d_n)-\pi_X\sigma\times\pi_Y\sigma]_{\#}\partial D(d_n)\\
&=(Id-\theta x)(\sigma)-\partial ( \pi_X\sigma_{\#} \otimes \pi_Y\sigma_{\#})D(d_n)\\
&=(Id-\theta x)\sigma-\partial D(\pi_X\sigma\times\pi_Y\sigma]_{\#}(d_n)\\
&=(Id-\theta x)\sigma-\partial  D\sigma\\
\end{split}$$
Note: the penultimate equality follows from our defintion of $D(\sigma)$.
