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.

 A: 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.
A: $U_R$ 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)
