I'm reading MacLane homology and topological Hochschild homology by Pirashvili-Waldhausen, and they define $L_0^{st}T$ for $T$ a functor $\mathbb P(R)\to R-\mathbf{Mod}$ ($R$ a ring, $\mathbb P(R)$ the category of finitely generated projective left $R$-modules) as follows :
extend $T$ so $s\mathbb P(R)$ (simplicial objects of $\mathbb P(R)$) degreewise, and then put $L_0^{st}T(P):=\pi_n T(K(P,n)), n>0$.
They refer to Dold-Puppe Homologie nicht additiver funktoren for the fact that $L_0^{st}T$ is additive, but in that paper they don't actually prove the result, they merely say "Man kann zeigen, dass $L_0^{st}T$ stets additiv ist" ("one can show that $L_0^{st}T$ is always additive").
I'd like to see a proof of that fact or a reference that actually proves it.
What I've tried to do :
First of all, it's easy to see that one may assume that $T(0)=0$, which might simplify things.
Then using the explicit model for $K(P,n)$ that comes from the Dold-Kan correspondance applied to the chain compex $P[n]$, we get that our $\pi_n$ is nothing but the quotient of $T(P)$ by $T(P^{n+1})$ where the map is $\sum_i (-1)^i T(d_i)$, where $d_i : P^{n+1}\to P$ is the face of the given model of $K(P,n)$ (note that $\sum_i (-1)^id_i = 0$)
So if we give ourselves a decomposition $P\cong Q\oplus K$ (which yields $P^{n+1}\cong Q^{n+1}\oplus K^{n+1}$) we get that we want to show that the induced map between the cokernels in the following square is an isomorphism :
$\require{AMScd}\begin{CD} T(P^{n+1})@>>> T(Q^{n+1})\oplus T(K^{n+1}) \\ @VVV @VVV \\ T(P)@>>> T(Q)\oplus T(K)\end{CD}$
It's easy to see that the bottom horizontal map is an epimorphism, so it suffices to show that the map between cokernels is an injection. To say that it is an injection amounts to (if I'm not mistaken) proving that the induced map $cr(Q^{n+1},K^{n+1})\to cr(Q,K)$ is an epimorphism, where $cr(A,B)$ is the cross-effect, that is the kernel of $T(A\oplus B)\to T(A)\oplus T(B)$, but I can't see any justification for it (that map is $\sum_i (-1)^i cr(d_i,d_i)$)
Maybe there's a more conceptual approach ?
Maybe one could use the fact that it doesn't depend on $n$ (which is actually proved in the relevant literature) and just do it for $n=1$ : for $n=1$ the $d_i$'s are the two projections (for $i=0,2$) and the sum (for $i=1$) so that the differential is $T(\pi_1)+T(\pi_2)-T(\pi_1+\pi_2)$ ($\pi_i$ the $i$th projection) but I can't see how this helps.
I found the answer basing myself on a remark I found in some other paper. The remark is :
The proof of that is pretty basic (it relies on the fact that if $f,g: A\to B$ then $f+g$ can be seen as the composite $A\overset{(f,g)}\to B\oplus B \overset{p_1+p_2}\to B$.
Using that, we note that in the case $n=1$, we have a natural (in $T$ and in $P$) exact sequence $T(P^2)\to T(P)\to L_0^{st}T(P)\to 0$, where the first map is precisely $T(p_1)+T(p_2)-T(p_1+p_2)$.
In particular, we have a natural epimorphism $T\to L_0^{st}T$, and we can examine the following commutative square :
$\require{AMScd}\begin{CD}T(P^2)@>>> L_0^{st}T(P^2) \\ @VVV @VVV \\ T(P) @>>> L_0^{st}T(P)\end{CD}$
where the vertical maps are $\bullet(p_1) + \bullet(p_2) - \bullet(p_1+p_2)$
Going down-right gives $0$ by definition of our natural epimorphism, so going right-down must be $0$ too. It follows (because right is an epimorphism) that $L_0^{st}T(P^2)\to L_0^{st}T(P)$ is $0$ as well, from which we get, by the lemma, that $L_0^{st}T$ is additive.