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On page 16 of his book on symmetric spectra, Stefan Schwede defines the $n$-th Eilenberg-Mac Lane space $(HA)_n$ of an abelian group $A$ by means of a construction called the reduced $A$-linearization. Given a pointed space $K$, the reduced $A$-linearization $A[K]$ consists of all linear combinations of points in $K$ with coefficients in $A$, where all multiples of the basepoint are identified with $0$. This is then topologized via the obvious surjection coming from $\bigsqcup_{n\geq 0} A^n \times K^n$.

Suppose that $x =\sum_i a_i f_i$ is a singular $n$-chain in $K$ with coefficients in $A$. This then defines a map $\hat{x} \colon \Delta^n \to A[K]$ by adding the $f_i$ pointwise and multiplying with the coefficients. Schwede claims that if $x$ is a cycle, the map $\hat{x}$ sends $\partial \Delta^n$ to $0$. This would then give a map $\widetilde{H}_*(K; A) \to \pi_*(A[K])$. However, I don't see why $\hat{x}|_{\partial \Delta^n} = 0$. Suppose for instance that $A = \mathbf{Z}/2$ and $K = \Delta^2$. Then the sum of the three edges gives a 1-cycle $x$ on $\Delta^2$, hence a map $\hat{x} \colon \Delta^1 \to \mathbf{Z}/2[\Delta^2]$. But $\hat{x}(0)$ is the sum of the three vertices of $\Delta^2$, which is not $0$. What am I doing wrong?

merle
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    I think you're right and there's a mistake in Schwede's book. What he's trying (I think) to say is that if you have a simplicial abelian group $A$ and you take its homotopy groups, then you recover the homology groups of the associated chain complex. This much is true, and in fact there's an easy to define map $\pi_(A)\to H_(\tilde A)$. But to build the map in the other direction, even in the simplicial setting (which is surely easier to handle than the topological one), you first have to normalize. That is, either consider a subcomplex or a quotient complex – Maxime Ramzi Jul 24 '20 at 16:07

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