For concreteness, let us work with the language of spectra introduced in EKMM.


In Strickland's paper "Products on $MU$-modules", he proves the following. If $R$ is a q-cofibrant commutative $S$-algebra which is even, in the sense that its homotopy is concentrated in even degrees, and $x\in R\_d$ is a non-zero divisor, then the diagram


$R/x \vee R/x \to R/x\wedge R/x \to \Sigma^{2d+2}R$


induces a left exact sequence when applying $[-,R/x]$. This is lemma 3.6 in his paper.


Since $\Sigma^{d+1} R$ is the cofiber of the canonical map $\rho:R\to R/x$, this led me to consider the following more general question:



> 
> Let $A$ be an $R$-algebra. Denote by $\bar{A}$ the cofiber of the unit map $R\to A$. Is there a cofiber sequence $A\vee A \to A \wedge A \to \bar{A}\wedge \bar{A}$ ?
> 
> 
> 


I tried playing around with diagrams but I got nowhere. It also seems the question would make sense in more generality, like in a stable monoidal model category with suitable additional hypotheses.



Cofiber sequence $A\vee A \to A \wedge A \to \bar{A}\wedge \bar{A}$ for a spectrum $A$
This will not work in exactly the form that you state, as you will see if you consider the case where $A=S\vee\overline{A}$; there is an extra factor of $S$. To fix this, you should replace $A\vee A$ by the pushout of the diagram $A\xleftarrow{}S\xrightarrow{}A$, which we can call $P$. It is then true that the cofibre of the natural map $P\to A\wedge A$ is $\overline{A}\wedge\overline{A}$. The easiest way to see this is to use the theory of total cofibres: given a commutative square
$$ \begin{array}{ccc}
 W & \xrightarrow{f} & X \\
 g \downarrow && \downarrow h \\
 Y & \xrightarrow{k} & Z
 \end{array}
$$ 
we can form the pushout $P$ of $f$ and $g$, then there are natural equivalences
$$ \text{cof}(P\to Z) \simeq\text{cof}(\text{cof}(f)\to\text{cof}(k)) \simeq 
 \text{cof}(\text{cof}(g)\to\text{cof}(h)).
$$
Apply this to the square 
$$ \begin{array}{ccc}
 S & \to & A \\ \downarrow && \downarrow \\ A & \to & A\wedge A
 \end{array}
$$
to get the earlier claim.


