Lemma 16.3.4 (Smash product). Let \(C\) be a symmetric monoidal \(\infty \)-category that admits finite colimits and a terminal object, and assume that \(X \otimes -\colon C\to C\) preserves finite colimits for every \(X\in C\). Equip \(\Ar (C)\) with the pushout-product monoidal structure of Lemma 16.3.3. Then \(C_*\subseteq \Ar (C)\) is symmetric monoidal, and the localization functor \[ \cofib \colon \Ar (C)\to C_*, \qquad (f\colon X\to Y)\longmapsto (*\to \cofib (f)) \] is symmetric monoidal. The tensor product of pointed objects is therefore \[ (x\colon *\to X)\otimes (y\colon *\to Y) = \cofib (X\otimes *\sqcup _{*\otimes *}*\otimes Y\to X\otimes Y). \]
Proof. The inclusion \(C_* \hookrightarrow \Ar (C)\) admits a left adjoint given by taking cofibers. By Proposition 14.5.3, it thus remains to show that tensoring in \(\Ar (C)\) with an object preserves those morphisms (= commutative squares in \(C\)) which induce isomorphisms on cofibers. To see this, consider such a morphism
Let us write \(Z := \cofib (f\colon X \to Y)\) and \(Z' := \cofib (f'\colon X' \to Y')\), so that the condition on the square is that the induced map \(Z \to Z'\) is an isomorphism. Our task is to show that for every other morphism \((g\colon A \to B)\) in \(C\) the induced commutative square
again induces an isomorphism on (vertical) cofibers. Since \(Z \iso Z'\), it will suffice to express the two vertical cofibers purely in terms of \(g\), \(Z\) and \(Z'\). We will do so by showing that the bottom front square in the following commutative diagram is a pushout square:
But this follows from the pasting law for pushout squares: the left and right faces of the diagram are pushouts by definition, while the back square and front rectangle are pushout squares since \(- \otimes A\) and \(- \otimes B\) preserve the pushout square defining \(Z\) as a cofiber of \(X \to Y\). □
Generated from the authoritative LaTeX source.