# Extend Alexander-Whitney and Eilenberg-Zilber map to n-fold tensor products

See the definition of the Alexander-Whitney transformation:

http://ncatlab.org/nlab/show/Alexander-Whitney+map

and the Eilenberg-Zilber transformation:

http://ncatlab.org/nlab/show/Eilenberg-Zilber+map

Is there a natural or obvious way to extend them to higher tensor powers i.e to, lets say

$$\Delta_{A_1,\ldots,A_n} : C(\otimes_{j=1}^n A_j) \to \otimes_{j=1}^n C(A_j)$$ and $$\nabla_{A_1,\ldots,A_n} : \otimes_{j=1}^n C(A_j) \to C(\otimes_{j=1}^nA_j)$$

or to the infinite tensor power series, such that the adjointness is still there?

(My first obvious guess is to simply iterate them using associativity of the usual tensor product but I'm not sure if it is that simple due to concerns about braiding and singns)

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Your obvious guess works. There are not more problems in the $n$-fold case than in the $2$-fold case. –  Fernando Muro Feb 11 '12 at 0:20
Ok. Why is that? The Alexander-Whitney transformation is defined in terms of simplicial sets by injections of $\lambda_{low}: \Delta_p \rightarrow \Delta_n$ and $\lambda_{high}: \Delta_q \rightarrow \Delta_n$, such that the first is injected to the lowest face and the second is injected to the highest face. But iterating this depends on the order of factors, doesn't it? The question is, how this extended to more than two factors accordingly? How to inject $n$ simplices $\Delta_{p_j} \rightarrow \Delta_k$ (with $\sum p_j = k$) propperly? The Eilenberg-Zilber map on the other and depends on –  Nevermind Feb 11 '12 at 0:37
the shuffel and I don't know how to extend its definition in a non order depended way to more than two factors –  Nevermind Feb 11 '12 at 0:38

Yes, there is a generalization to a finite number of simplicial complexes. A reference is Corollary 2.2 in the paper

Eilenberg, MacLane: On the groups $H(\Pi,n)$, II: Methods of Computation, Ann. of. Math. 60(1954), No. 1, 49 - 139.

Using the definitions from nLab, the maps are given as follows:

1) Let $a_i \in A_i$ be homogen. $$\nabla(a_1 \otimes \cdots \otimes a_n) = a_1 \nabla a_2 \nabla ... \nabla a_n$$ (well-definied since $\nabla$ is associative)

2) Let $a_i \in (A_i)_m$. $$\Delta(a_1 \otimes \cdots \otimes a_n) = \sum \displaystyle \otimes_{i=1}^n\displaystyle\tilde{d}^{m-j_i}d_0^{j_{i-1}}a_i$$ where the sum is taken over $0 \le j_1 \le \cdots \le j_{n-1} \le m$ and $\tilde{d}^{m-j_n},\;d_0^{j_0}$ has to be interpreted as identity.

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oh nice. Is the equation in the paper you mentioned? –  Nevermind Feb 11 '12 at 0:59
Yes. (the editor requires more words for a comment, so I'll take the opportunity to welcome you at Mathoverflow :). –  Ralph Feb 11 '12 at 1:06