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Let $X, Y$, let's say, homotopy commutative $H$-spaces, $f,g$ maps from $X$ to $Y$. (Actually we only need $Y$ to be homotopy commutative $H$_space, but the statement is easier if we also suppose $X$ to be one). It is not difficult to show that

$Q(f+g)_*=Qf_* +Qg_*$ where $Q$ denotes the module of indecomposables, and $_*$ denotes, let's say, induced map in mod $p$ ordinary homology. (Actually any generalized homology works here as long as we have Kunneth isomorphism.)

This can be shown either by directly writing down the two sides (if we denote $$\Delta (x)= x \otimes 1 +1\otimes x +\Sigma x' \otimes x''$$ then, we have $$(f+g)_*(x)= f_*x +g_*x + \Sigma f_*x' \otimes g_*x''\mbox{ })$$ or by noting that both $_*$ and Q are additive functors.

Now my question is: I have seen this written somewhere, could anyone tell me where I can find a published reference?

This is something completely trivial once you know. However,

  1. it is not so well-known
  2. if I hadn't read it some where, then I probably wouldn't have found it on my own
  3. it makes the life a lot easier.

To illustrate these points, I can say that I have seen a paper where the author computes $(f+g)_*$ in a great length then only to pass to the indecomposable quotients. So the author and the referee of the paper weren't able to come up with this.

So I really would like to give credit to whoever wrote this fact in a published paper. Thank you in advance.

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  • $\begingroup$ Is this not immediate from working with the dual problem and noting that $Q(f+g)_*$ is dual to $Pf^*$, $Pg^*$, where $P$ is the module of primitives in the mod p cohomology? $\endgroup$
    – Tyrone
    Jan 27, 2018 at 9:39
  • $\begingroup$ @Tyrone Well, I am not sure considering only elements for which there is no "extra terms" in the diagonal and then dualizing is more "immediate" than quotienting out tje extra terms in the diagonal. $\endgroup$
    – user43326
    Feb 2, 2018 at 18:25
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    $\begingroup$ Note that, since the definition of $f+g$ uses the diagonal and the multiplication in symmetric ways, that the same argument you give also says that $(f+g)_* = f_* + g_*$ on homology primitives. I used this fact, without comment, in my 2015 J.Topology paper (see Theorem 1.4). I am sure many people have similarly used this. If you want a reference, maybe search in Joe Neisendorfer's book, as folks who study finite loop spaces and their decompositions need these sorts of tools. $\endgroup$
    – Nicholas Kuhn
    May 19, 2018 at 2:30
  • $\begingroup$ Thank you very much Nick. Unfortunately I couldn't find it in Neisendorfer's book or his papers. $\endgroup$
    – user43326
    Oct 3, 2018 at 9:24

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