Diagonal maps, Goodwillie calculus, and $T(n)$ local homotopy theory Here is a collection of facts that all seem true, but together seem to give a nonsensical solution:

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*After $T(n)$-localization, all natural transformations $F \sim G$ between homogenous functors $F,G:Sp \rightarrow Sp$ of different degrees appear trivial, in the sense that $\operatorname{cofiber}(\sim) \simeq G \vee\Sigma F$. This follows from Kuhn's splitting.


*If one has compositions of functors $f \circ g, f' \circ g'$ and natural transformations $\alpha: f \rightarrow f',\beta: g \rightarrow g'$ such that either $\alpha$ or $\beta$ is trivial on derivatives, then $\alpha \circ \beta$ is trivial on derivatives. This follows from the chain rule of Arone and Ching that the derivatives can be computed as a (derived) composition product.


*The natural transformation $\Sigma^\infty \Omega^\infty \rightarrow \Sigma^\infty (\Omega^\infty(-))^{\wedge 2}$ given by $1 \circ \Delta \circ 1$ induced by the diagonal $\Delta$ in $Top_*$ is $0$ on derivatives. This is because $\Sigma^\infty \Delta: \Sigma^\infty \rightarrow \Sigma^\infty(-)^{\wedge 2}$ is a natural transformation between homogenous functors of different degrees.


*The Taylor tower $(\Sigma^\infty \Omega^\infty )^{\wedge n}$ converges on $0$-connected spectra. I believe this follows from the fact for $n=1$.


*Hence, for a connected infinite loop space $X$, the cofiber of $\Sigma ^\infty \Delta: \Sigma^ \infty X \rightarrow \Sigma^\infty (X \wedge X)$ is $T(n)$-locally equivalent to $\Sigma^\infty (X \wedge X) \vee \Sigma^\infty X$.
This seems to imply that $T(n)$-locally, the diagonal maps of infinite loop spaces appear trivial in that they can't be detected by the cofiber.
I am wondering if this conclusion is correct, or if I have an error. I originally asked this question rationally, but figured I'd ask it in the $T(n)$-local category since it also applies there.
 A: This is an elaboration on my comment above. Let us consider the natural transformation induced by the diagonal
$$\Sigma^\infty\Delta\colon \Sigma^\infty X \to \Sigma^\infty X\wedge X.$$
The natural transformation $\Sigma^\infty \Delta$ induces a map between the Goodwillie derivatives of the functors. Let us denote this induced map by
$\partial_*\Sigma^\infty\Delta\colon \partial_* \Sigma^\infty\to \partial_* \Sigma^\infty \mbox{Sq}$. Here Sq denotes the functor Sq$(X)=X\wedge X$. The question is what is $\partial_*\Sigma^\infty\Delta$, and in particular if it can be non-trivial.
The functor $\Sigma^\infty$ is homogeneous linear, and $\Sigma^\infty$Sq is homogeneous quadratic. So we have the following description of the derivatives as symmetric sequences
$$\partial_*\Sigma^\infty = (\Sigma^\infty S^0, *, *, \ldots, )$$
$$\partial_* \Sigma^\infty\mbox{Sq} = (*, \Sigma^\infty{\Sigma_2}_+, *, *, \ldots, ).$$
The space of maps from $\partial_*\Sigma^\infty$ to $\partial_* \Sigma^\infty\mbox{Sq}$ in the $\infty$-category of symmetric sequences is easily seen to be contractible. So it seems that $\partial_*\Sigma^\infty \Delta$ can only be the trivial map. However, we really are interested in the mapping space in the category of right modules over $\partial_*Id$. Let's see what this means. Let $O=\{O(n)\}$ be an operad in spectra. We consider reduced operads, which means that $O(0)=*$ and $O(1)=\Sigma^\infty S^0$. All our modules are truncated at $2$, so only $O(1)$ and $O(2)$ are relevant. Let $M(1), M(2)$ and $N(1), N(2)$ be (truncated) right modules over $O$. This means that we have $\Sigma_2$-equivariant maps $M(1)\wedge O(2) \to M(2)$ and $N(1)\wedge O(2) \to N(2)$. The mapping spectrum of $O$-module maps from $M$ to $N$ is equivalent to the homotopy pullback of the following diagram
$\require{AMScd}$
$$\begin{CD}
 @. \mbox{Spectra}(M(2), N(2))^{h\Sigma_2}\\
@. @VVV\\
\mbox{Spectra}(M(1), N(1)) @>>> \mbox{Spectra}(M(1)\wedge O(2), N(2))^{h\Sigma_2}
\end{CD}$$
By contrast, the space of maps from $M$ to $N$ in the category of symmetric sequences is the product
$$\mbox{Spectra}(M(1), N(1))\times \mbox{Spectra}(M(2), N(2))^{h\Sigma_2}.$$
There is an obvious forgetful map from the former mapping spectrum to the latter, which sends the middle term to a point.
In the case of interest, $O=\partial_*Id$, so $O(1)=\Sigma^\infty S^0$ and $O(2)\simeq\Sigma^{-1}\Sigma^\infty S^0$, where $\Sigma_2$ acts trivially on $O(2)$. The modules are $(M(1), M(2))=(S^0, *)$ and $(N(1), N(2)) = (*, \Sigma^\infty {\Sigma_2}_+)$. So the spectrum of $O$-module maps from $M$ to $N$ is the following pullback
$$\mbox{Spectra}(\Sigma^\infty S^0, *)\to \mbox{Spectra}(\Sigma^{-1}\Sigma^\infty S^0, \Sigma^\infty {\Sigma_2}_+)^{h\Sigma_2} \leftarrow \mbox{Spectra}(*, \Sigma^\infty{\Sigma_2}_+)^{h\Sigma_2}.$$
This homotopy pullback is easily seen to be equivalent to $\Sigma^\infty S^0$.
It remains to show that the transformation $\Sigma^\infty\Delta$ actually induces a non-trivial map on derivatives. Here is one way to argue this. It is an exercise in Yoneda that the spectrum of natural transformations from $\Sigma^\infty X$ to $\Sigma^\infty X\wedge X$ is equivalent to the sphere spectrum, with the diagonal corresponding to a generator. There is a natural map of mapping spectra $$\mbox{Nat}(\Sigma^\infty X, \Sigma^\infty X\wedge X)\to \partial_*Id\!-\! mod(\partial_*\Sigma^\infty X, \partial_*\Sigma^\infty X\wedge X).$$ Both the source and the target are equivalent to $\Sigma^\infty S^0$. One can show that for polynomial functors whose derivatives are finite free spectra (or more generally, whose derivatives have vanishing Tate homology) this map is an equivalence.
