# Forgetting and tensoring up for very connective maps of $E_{\infty}$-rings

Does anything happen if I forget and tensor back up along a highly connective map of $E_{\infty}$-rings?

Here's what I mean precisely: Let $f \colon A \to B$ be a $n$-connective map between connective $E_{\infty}$-rings, with $n \geq 1$. Here $n$-connective means that $\pi_i (fib(f))=0$ for $i < n$. Roughly, $f$ is a very surjective map, and if we let $fib(f)=I$, then $B$ is roughly $A/I$.

Now let's take a $B$-module $M$. We can forget the $B$-module structure and view it as an $A$-module. Then we can tensor it back up again to get $B \otimes _A M$. By adjunction there is a canonical map $m \colon B \otimes _A M \to M$, which is the multiplication map. How far is this map from being an equivalence?

The case I am really interested is when $f \colon A \to B$ is a square-zero extension obtained from a $n$-connective derivation $\eta \colon L_B \to M[1]$. Then $fib(f)$ can be identified with $M$, and so additonally has the structure of a $B$-module. Does anything special happen in this case?

(Derived) tensoring preserves fiber sequences. In particular, there is a fiber sequence $$fib(f) \otimes_A M \to M \to B \otimes_A M.$$ If you allow additional assumptions, you can say more about this. For instance, if $A$ and $B$ are connective, then the Kunneth/hypertor spectral sequence $$Tor^{\pi_*A} (\pi_* fib(f), \pi_* M) \Rightarrow \pi_*(fib(f) \otimes_A M)$$ will tell you that the connectivity of the fiber is the sum of the connectivity of $M$ and the connectivity of $f$ (in the terminology that you're using). Roughly, then, you find that the map $M \to B \otimes_A M$ is an isomorphism on the same number of homotopy groups that $A \to B$ is.
In the special case where $fib(f)$ is $M$ itself, you're getting a fiber sequence $$M \otimes_A M \to M \to B \otimes_A M.$$ However, the map on the left is the multiplication map, which is zero by the square-zero assumption, and so this degenerates to an equivalence $$B \otimes_A M \simeq M \vee (M \otimes_A M)[1].$$
• I'm still puzzled by something. In the first part of your argument (not the square-zero part) it is never used that $M$ has the structure of a $B$-module, and the argument is in fact valid for any $A$-module. In the case of discrete rings, when I can really identify $B$ with $A/I$, your first fiber sequence becomes $IM \to M \to M/IM$. If $M$ is an $A/I$-module, $IM$ is actually zero. Does nothing remotely similar hold in homotopical algebra? Mar 14, 2012 at 10:12
• Ah! I missed the B-module portion of the hypothesis. In that case, you can in fact use the module structure to split the cofiber sequence I listed above, and find a similar identity: $B \otimes_A M \simeq M \vee fib(f) \otimes_A M[1]$. Is that along the lines of what you were thinking? (In the case you listed, you don't actually have $IM = 0$ in the derived setting, because you might have elements representing higher relations.) Mar 14, 2012 at 14:12