Operadic Lift of Lurie's Relative Tensor Product

In Section 4.4 of his book Higher Algebra, Lurie introduces, for a monoid object $A$ of a monoidal quasicategory $C$, and right and left $A$-modules $M,N$, the relative tensor product $M\otimes_AN$. As I've stated it, this object is only a $(1,1)$-bimodule, where I'm using $1$ to indicate the monoidal unit of $C$. If $M$ and $N$ have left and right $A'$ and $A''$-actions respectively, then $M\otimes_A N$ is an $(A',A'')$-bimodule.

In general, this construction (see Construction 4.4.2.7) is done over an arbitrary $\infty$-operad $O^\otimes$. This uses Lurie's $\infty$-operad $Tens^\otimes$ in an essential way. Specifically if we have a commutative diagram

$$\matrix{ Tens^\otimes_{[2]} \to& C^\otimes\\ \downarrow & \downarrow\\ Tens^\otimes\overset{f}\to& O^\otimes}$$

where the top horizontal map picks out the pair of modules and the algebra we wish to tensor them over, then the relative tensor product is given by a lift $Tens^\otimes\to C^\otimes$ as a diagonal arrow in this diagram, whose restriction to a certain subcategory of $Tens^\otimes$ picks out the object in $C$ which is the tensor product itself.

My Question is the following: What sorts of maps are there $f:Tens^\otimes\to O^\otimes$? For instance, since $Tens^\otimes$ is produced, it seems, from $N(\Delta)^{op}$, I cannot see how one might get a map $f:Tens^\otimes\to E_n^\otimes$ to produce the relative tensor product over an $E_{n-1}$-algebra in $C$. Is there a way around this? Does Lurie describe how to produce more than maps $Tens^\otimes\to Ass^\otimes$?

• In the diagram you want the object at the bottom left corner to be $Tens^\otimes_\rightY$, not just $Tens^\otimes$. My feeling on this is that in order to take the relative tensor product of $E_n$-algebras, you need to choose a way to regard them as associative algebras (otherwise you don't know how to "line up" the factors). This is the same as choosing a map $Ass^\otimes \to E_n^\otimes$. – Saul Glasman Nov 18 '15 at 19:24
• @SaulGlasman Yeah, I just didn't know the symbol, so I just called it $Tens^\otimes$ – Jonathan Beardsley Nov 18 '15 at 19:33
• Okay, and so we use the composition $Tens^\otimes_{\rightY}\to Ass^\otimes\to E_n^\otimes$? – Jonathan Beardsley Nov 18 '15 at 19:34
• @SaulGlasman Interesting. Does this issue not appear in the commutative case or is it that we have some canonical way of viewing in $E_{\infty}$-algebra as an $E_1$-algebra that does not generalize to the $E_n$-algebra case? – Sean Tilson Nov 19 '15 at 12:31