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$\DeclareMathOperator{\Sp}{\mathrm{Sp}}$I am taking a special case $\Sp$ here, mainly because it has nice categorical properties.

Let $R$ be an $E_\infty$-ring spectrum. In Higher Algebra, Lurie proves we have a forgetful functor (part of monadic adjunction) $$ U_R:\operatorname{Mod}_R(\Sp) \rightarrow \Sp$$ where $\Sp$ is in the $\infty$-category of spectra.

$U_R$ reflects equivalences. But is $U_R$ faithful in the sense that that the induced map of $$Map(x,y)\rightarrow Map(U_Rx,U_Ry)$$ mapping spaces is $-1$-truncated in the $\infty$-category of spaces. i.e. the homotopy fibers are $-1$-truncated.

One categorically, $U$ is faithful in many cases, i.e. if we replace $\Sp$ with $\mathrm{Ab}$. Perhaps the answer is false in $\infty$-categories. I'd like to understand what goes wrong. Some comments on the following would be helpful:

  • A counter example where $U_R$ is not faithful. (i.e. is it faithful when $R=H\Bbb Z$? )
  • A brief/reference explanation for what accounts of this.
W. Zhan
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    There isn't even a good notion of faithfullness for $\infty$-categories. A good replacement is to ask for maps on mapping spaces to be injective on $\pi_0$ and isomorphisms on all higher $\pi_i$. But that's basically never satisfied in this case. – Achim Krause Oct 25 '20 at 08:13
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    @AchimKrause I thought that was the notion of faithfulness for $\infty$-categories, a functor which is a homotopy monomorphism on hom-spaces. – Alexander Campbell Oct 25 '20 at 08:19
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    Ok, yeah, maybe that's a better way to put it. There IS such a notion, but it's far too strong since it requires isomorphisms on all higher homotopy groups of mapping spaces. – Achim Krause Oct 25 '20 at 08:21
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    Ok, so I added my understanding of mono. @Achim, may you explain a little why is this notion of mono almost never satisfied in this case? – W. Zhan Oct 25 '20 at 08:46
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    The two answers given explain nicely what goes wrong in specific examples. I want to add the observation that an exact functor between stable $\infty$ categories is faithful if and only if it is actually fully faithful. In your case, this happens precisely if $R$ is a spectrum with $R\otimes_{\mathbb{S}} R = R$, i.e. some kind of localisation. – Achim Krause Oct 25 '20 at 12:48
  • @AlexanderCampbell, in the case of an adjunction, the right adjoint is faithful iff the counit is an epi. Is there any chance that one can you a variation of this concept, at least to define a notion of faithfulness for a right adjoint? – Ivan Di Liberti Oct 25 '20 at 21:00
  • @IvanDiLiberti This continues to hold for $\infty$-categories, where faithful is defined as above, and a morphism $f \colon A \to B$ in an $\infty$-category $\mathcal{C}$ is an epimorphism if the pre-composition map $\mathcal{C}(f,C) \colon \mathcal{C}(B,C) \to \mathcal{C}(A,C)$ is a (homotopy) monomorphism of $\infty$-groupoids for every object $C \in \mathcal{C}$. – Alexander Campbell Oct 26 '20 at 00:48

2 Answers2

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$U_R$ obviously preserves delooping, so if that were the case, because $\pi_0 map(X,Y) = \pi_1 map(X, \Sigma Y)$, you would also get an isomorphism on $\pi_0$, so an equivalence of mapping spaces.

In other words, $U_R$ is faithful if and only if it is fully faithful. But now for a map of ring spectra $R\to S$, the forgetful $Mod_S \to Mod_R$ is fully faithful if and only if $R\to S$ is an epimorphism of ring spectra (good examples are localizations - be careful that classical examples such as $R\to R/I$ for a usual ring $R$ tend to fail).

This is to say that "being an $S$-module" becomes a property of an $R$-module, rather than additional structure - so of course you can expect that to be very rare.

In your example of $H\mathbb Z$, it doesn't hold at all - you can for instance detect it on the level of the ring of stable cohomology operations of singular cohomology, which is bigger than just $\mathbb Z$ (look at the (co)homology of Eilenberg-MacLane spaces)

Maxime Ramzi
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  • Thanks a lot Maxime. For the second paragraph, is there a reference of this statement/ what do you mean by epi of ring spectra? – W. Zhan Oct 30 '20 at 07:32
  • I mean one of the following 2 equivalent statements : 1- $R\otimes_S R\to R$ is an equivalence; 2- $map(R,A)\to map(S,A)$ is a monomorphism of spaces for all commutative ring spectra $A$. I briefly outlined a proof of their equivalence (and the equivalence with $Mod_R\to Mod_S$ being fully faithful) in my question here : https://mathoverflow.net/questions/370081/interesting-epimorphisms-of-e-infty-ring-spectra . As for a reference, I'm not sure - you can try to see Higher Topos Theory section 5.2.7. about the relationship between fully faithfulness of the right adjoint and other – Maxime Ramzi Oct 30 '20 at 08:48
  • conditions on the adjunction. This will tell you that the forgetful functor is fully faithful if and only if the co-unit $R\otimes_S M\to M$ is an equivalence ($M$ lives in $R$-modules). But both sides preserve colimits so this is an equivalence if and only if it is one when $R=M$, so you can compare fully faithfulness and the fact that $R\otimes_S R\to R$ is an equivalence. Then (cf. my question, but of course this is not original - I couldn't tell you where to find a specific reference) you can compare this condition and the epimorphism condition – Maxime Ramzi Oct 30 '20 at 08:50
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In general, the functor $U_R$ does not induce isomorphisms on higher homotopy groups of mapping spaces. Let $R=H(\mathbf{Z}/2)$. Then $\pi_*(map(R,R))$ is the Steenrod algebra $\mathcal{A}^*$ where $map$ denotes the mapping spectrum. The mapping spectrum $map(R,R)$ therefore has non-zero homotopy groups in negative degrees and differs from the mapping spectrum of $R$-module maps from $R$ to itself, which is just $R$ again, whose homotopy groups consist of $\mathbf{Z}/2$ concentrated in degree zero.

To see this difference directly in terms of mapping spaces as opposed to mapping spectra, we consider maps from $R$ to deloopings of $R$. For example, $$\pi_1(Map_{R-Mod}(R, R[2])) \cong \pi_0(Map_{R-Mod}(R, \Omega R[2])) \cong \pi_0(Map_{R-Mod}(R, R[1])) \cong \mathrm{Ext}^1_R(R,R) = 0$$ but $$\pi_1(Map_{Sp}(R,R[2])) \cong \pi_0(Map_{Sp}(R, \Omega R[2])) \cong \pi_0(Map_{Sp}(R,R[1])) = \mathcal{A}^1 \cong \mathbf{Z}/2$$ so the induced map on $\pi_1$ is not surjective.

Daniel Bruegmann
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