# Is the forgetful functor $\mathrm{Mod}_R \mathrm{Sp} \rightarrow \mathrm{Sp}$ faithful?

$$\DeclareMathOperator{\Sp}{\mathrm{Sp}}$$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.
• 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. Oct 25 '20 at 8:13
• @AchimKrause I thought that was the notion of faithfulness for $\infty$-categories, a functor which is a homotopy monomorphism on hom-spaces. Oct 25 '20 at 8:19
• 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. Oct 25 '20 at 8:21
• 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? Oct 25 '20 at 8:46
• 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. Oct 25 '20 at 12:48

$$U_R$$ 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)

• Thanks a lot Maxime. For the second paragraph, is there a reference of this statement/ what do you mean by epi of ring spectra? Oct 30 '20 at 7:32
• 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 : mathoverflow.net/questions/370081/… . 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 Oct 30 '20 at 8:48
• 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 Oct 30 '20 at 8:50

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.