$\DeclareMathOperator\Hom{Hom}$It is well-known (see [Breen, Mikhailov, Touzé - Derived functors of the divided power functors](https://arxiv.org/abs/1312.5676) for example) that for $A$ a free abelian group we have
$$ H\_i(K(A,1); \mathbb{Z}) \cong {\bigwedge}^i A$$


the exterior powers of $A$
and that for $B$ an abelian group we have


$$ H\_2(K(B,2); \mathbb{Z}) = B, \quad H\_3(K(B,2); \mathbb{Z}) = 0, \quad H\_4(K(B,2); \mathbb{Z}) = \Gamma(B)$$


where $\Gamma(B)$ is Whitehead's gamma group. It is the universal recipient of a quadratic map. It can be defined as the free group generated by the symbols $\gamma(b)$ for $b \in B$ subject to two relations:


1. $\gamma(-b) = \gamma(b)$;
2. $\gamma(x + y + z) + \gamma(x) + \gamma(y)+ \gamma(z) = \gamma(x+y) + \gamma(y+z) + \gamma(z+x)$.


A map $f:K(A, 1) \to K(B,2)$ is given by a degree 2 cohomology class. We can use universal coefficients to compute this, and since $A$ is free the ext term vanishes, giving us:
$H^2(K(A,1); B) = \Hom( \wedge^2 A, B)$.


Thus given $f: {\bigwedge}^2 A \to B$, we have a map $f:K(A, 1) \to K(B,2)$, and thus an induced map on degree 4 homology: $f\_\*: {\bigwedge}^4 A \to \Gamma(B)$.


**Question**: Given $f$, what is this map $f\_\*: {\bigwedge}^4 A \to \Gamma(B)$?


Another way to phrase this is if I have a quadratic function on $B$ and a homomorphism $f$, then I should be able to combine these into a linear function on ${\bigwedge}^4 A$. How to do this?



Induced map on $H_4$ of Eilenberg–MacLane spaces
Let me summarize comments above. Functoriality of the resulting map $f\_\*:H\_4(K(A,1),\mathbb{Z})\to H\_4(K(B,2),\mathbb{Z})$ translates into a request for the universal map $\wedge^4 A\to \Gamma(A)$ which in turn, being functorial in $A$, allows to guess the answer assuming $A$ is free and
$rk\ A=4$.


Note that $\Gamma(A)$ identifies with the second divided powers $\Gamma^2(A)$
under the map $\gamma(x)\to \frac{x^2}{2}=\gamma^2(x)$. Thus we can write the product in divided powers as $x\cdot y=\gamma(x+y)-\gamma(x)-\gamma(y)$.


Consider a map given by
$$a\wedge b \wedge c \wedge d\to a\wedge b\cdot c\wedge d - a\wedge c\cdot b\wedge d+a\wedge d\cdot b\wedge c$$
Under assumption $rk\ A=4$ it defines a functorial map $\wedge^4 A\to \Gamma(A)$ which is unique up
to a scalar. To ensure that the universal $f\_\*$ is given by the above formula, we have to check that the scalar is indeed equal to $1$. For a prime $p$ we have the reduction inclusion $H\_4(K(\wedge^2 A,2),\mathbb{Z})/p\to H\_4(K(\wedge^2 A,2),\mathbb{Z}/p)$, thus it is enough to consider similar maps
$\bar{f\_\*}:H\_4(K(A,1),\mathbb{Z}/p)\to H\_4(K(\wedge^2 A,2),\mathbb{Z}/p)$ for all primes $p$, which are compatible with $f\_\*$ under the reduction. Then, following
the definitions and Serre's description of EM-space cohomology $\mod p$, one can see that
its dual is given by the usual multiplication
$Sym^2(\wedge^2 (A/p)^\*)\overset{\wedge}{\to} \wedge^4 (A/p)^\*$. Dualizing again we see that our universal formula is simply the lifting of this comultiplication and there are no unexpected multipliers.


