ScalingStacks

Proof. By 3.2.3, if $\mathcal{P}\not\simeq\mathbb{E}_{0}$, then it is $\left(-1\right)$-connected and in particular $\mathcal{P}\left(2\right)\neq\varnothing$. We shall construct a section to $\underline{f_{A,B}}$ using any binary operation $\mu\in\mathcal{P}\left(2\right)$. Let $i_{A}\colon A\to A\sqcup B$ and $i_{B}\colon B\to A\sqcup B$ be the canonical maps of the coproduct. Define $s$ to be the composition of the following maps:

$$\underline{A}\otimes\underline{B}\xrightarrow{\underline{i_{A}}\otimes\underline{i_{B}}}\left(\underline{A\sqcup B}\right)\otimes\left(\underline{A\sqcup B}\right)\xrightarrow{\mu_{A\sqcup B}}\underline{A\sqcup B}.$$

Now, consider the following diagram in the homotopy category of $\mathcal{C}$:

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The upper square commutes since $f_{A,B}$ is a map of algebras. The upper triangle commutes since it is the tensor product of two triangles, which commute by the very definition of $f_{A,B}$. The lower square commutes by the definition of the algebra structure on $A\otimes B$ and the lower triangle also clearly commutes. The composition of the bottom diagonal map and the bottom right map is the identity, since the restriction of $\mu$ to the unit in one of the arguments is homotopic to the identity map of the other argument. The composition of the top diagonal map with the top right map is $s$. It follows that $\underline{f_{A,B}}\circ s\sim\operatorname{Id}_{\underline{A\otimes B}}$. ∎