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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.)

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Can someone give me a hint to prove that?

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1 Answer 1

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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 $D(\sigma)$.

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