Does derived tensor product preserve fiber sequence？

Question

In lemma 3.1.5 of this paper I read, there is a fiber sequence of the underlying spaces of simplicial commutative rings $A\stackrel{f}{\longrightarrow} A\rightarrow A/\!\!/f$. Here we define the "derived quotient' $A/\!\!/f$ by the following homotopy pushout diagram: ($f$ is a point in $A$) which we can also denote as $A\otimes^{L}_{Z[T] } Z[T]/(T)$.

I am wondering how we deduce the fiber sequence from the above definition. Here are my thoughts on this:

By the above definition, consider the exact sequence $Z[T]\stackrel{\cdot T}{\longrightarrow} Z[T] \rightarrow Z[T]/(T)$ ( which also is a fiber sequence), after the fully faithful embedding into the category of simplicial ring, which is a right adjoint to $\pi_0$ functor, is again a fiber sequence because right adjoint preserves pullbacks.

if the $\textbf{derived tensor product commutes with fiber sequence in SCRing}$, then from the above sequence we have $A\otimes_{Z[T]}^{L}Z[T]=A\stackrel{\cdot f}{\longrightarrow} A\otimes_{Z[T]}^{L}Z[T]=A\rightarrow A\otimes_{Z[T]}^{L}Z[T]/(T)=A/\!\!/(f)$, which is a fiber sequence in $SCRing$, and geometric realization preserve pullbacks, thus we get the fiber sequences of the underlying spaces.

Here I just take the model category structure on $SCRing=Fun(\triangle^{op},CRing)$ and work with the (homotopy) (co)limits, I understand in many cases we are dealing with the infinity category structure on $SCRing$ as discussed in this post, probably using the infinity structure will be more convenient for the proof of this fiber sequence, any thoughts or comments on this question are very welcome.

Remark: I believe this problem boils down to the commutativity of homotopy limits and colimits, more precisely, pullback of push out is again pullback, sadly I can’t use Mather’s first cubic here because it requires somehow two pullback diagrams.

Answer 1

I have a possible answer for this question finally, but I am not that sure.

First of all, the question is not that 'correct' because the sequence $Z[T]\stackrel{\cdot T}{\longrightarrow} Z[T]\rightarrow Z[T]/(T)$ does not lie in the category $CRing$ but in $Mod$ (This is obvious because the first arrow is not a ring morphism), neither does $A\rightarrow A\rightarrow \displaystyle A/\!\!/f$, which actually lies in infinity category $Mod(Z[T])$. So we should consider everything here in the stable $\infty$ category $Mod$.

In $Mod(Z[T])$, by tensor product with $A\otimes_{Z[T]}^{L}(-)$, we have again an exact sequence $A\otimes_{Z[T]}^{L} Z[T]\rightarrow A\otimes_{Z[T]}^{L} Z[T]\rightarrow A\otimes_{Z[T]}^{L} Z[T]/(T)$, it is both fiber and also cofiber sequence because we are in stable infinity category, and recall that what we need is actually $\textbf{a fiber sequence of underlying spaces}$ instead of some fiber sequence in $\textbf{SCRing}$, luckily, the infinite loop functor $\Omega^{\infty}(-)$ functor preserves all small limits, thus we get a fiber sequence in $\textbf{Spc}$.

It's worth mentioning that the derived tensor product indeed is constructed from a pushout diagram in $\textbf{SCRing}$, but it is not a pushout when we consider the underlying module structure in $\textbf{Mod}$. To see this, just take the category of modules over some field $k$, then pushout in the category of yourvector space is exactly the cartesian product whose dimension is different from tensor product. Instead, 'tensor product $\otimes$' usually inherits from some more basic underlying category, for example here in the $E_{\infty}$ context, the tensor product(or the symmetric monoidal structure) actually comes from the smash product in the infinity category of spectra $\textbf{Sp}$. So this is not a question about commutativity of limits and colimits because in $Mod$, tensor product is not a colimit at all.

More precisely, to fit in the definition we used at the beginning $\displaystyle A/\!\!/f=A\otimes_{Z[T]}Z[T]/(T)$ which is the pushout in $SCRing$, for all these 'rings', consider their underlying 'modules' and this pushout has the tensor product of 'modules' as its underlying 'module', thus they have the same underlying spectrum in $\textbf{Sp}$. This explains our abuse of notation of tensor product $\otimes$ here. Also see this relevant post.