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Let's will write $K_n$ for the Eilenberg-MacLane space $K(\mathbb{Z},n)$. I remind that $K_n$ is equivalent to the loop space of $K_{n+1}$.

Let’s consider the map $\smallsmile:K_n\times K_m \to K_{n+m}$ corresponding to the cup product.

Given two elements $x:K_n$ and $y:K_m$, we can see $x$ as a loop in $K_{n+1}$, then pair it with $y$ to obtain a loop in $K_{n+1}\times K_m$, apply the cup product there (we obtain a loop in $K_{n+m+1}$), and finally see that loop as an element of $K_{n+m}$. It turns out that what we obtain is equivalent to the cup product of $x$ and $y$ (there might be a sign depending on how things are set up).

Is there a reference for this result? It seems rather basic but I couldn’t find it discussed anywhere.

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It is a bit of a roundabout way of proving it, but what you're looking for is immediately implied by the fact that the Eilenberg-Mac Lane spectrum is a commutative ring spectrum, as proven for example in in Example 1.14 of Stefan Schwede's book on symmetric spectra (note that $A[S^n]$ in that section is an explicit model for a $K(A,n)$).

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  • $\begingroup$ Thanks! Do you know if it goes both ways? I mean, is what I’m looking for somehow an essential part of the fact that the Eilenberg-MacLane spectrum is a commutative ring spectrum ? (if that means something) $\endgroup$
    – Guillaume Brunerie
    Commented Apr 27, 2016 at 16:14
  • $\begingroup$ @GuillaumeBrunerie Yes, in fact it is basically part of the definition, although the condition you ask about is more related to associativity than commutativity (Note: I use associative and commutative ring spectrum as synonymous to $E_1$ and $E_\infty$-ring spectrum respectively. The usage you can find in old papers is different but hopefully it's disappearing by now) $\endgroup$
    – Denis Nardin
    Commented Apr 27, 2016 at 16:18

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