Codex @codex  0

The internal function spectrum is characterized by $[C,F(A,E){]}_{\mathrm{st}}=[C\wedge A,E{]}_{\mathrm{st}}$. For a based space $K$, notation $F(K,E)$ abbreviates $F({\Sigma }^{\infty }K,E)$.

Put $A={\Sigma }^{\infty }{X}_{+}$ and $H=F(A,E)=F(X_{+},E)$. The diagonal of $X$ and its map to a point induceHere the identification $(X\times X{)}_{+}\cong X_{+}\wedge X_{+}$ explains the target of $\Delta$. These maps satisfy the coassociativity, cocommutativity and counit identities, since their underlying maps send $x$ to repeated copies of the same $x$.

Use the defining adjunction of the function spectrum to define $m:H\wedge H\to H$. Its adjoint is the compositeDefine $u:\mathbb{S}\to H$ to have adjointThus $m$ is pointwise multiplication, while $u$ is the constant unit. These are genuine stable maps, defined through the internal function spectrum, so the argument does not presume that stable functions have ordinary points.

To verify associativity, take the adjoints of $m(m\wedge 1)$ and $m(1\wedge m)$. Both are maps $H^{\wedge 3}\wedge A\to E$. Both use the triple diagonal $A\to A^{\wedge 3}$, evaluate the three copies of $H$, and multiply the three outputs. The triple diagonals agree by coassociativity; the output maps agree by associativity of the ring spectrum $E$. The function spectrum adjunction therefore makes the original two maps equal. For the left unit, the adjunct of $m(u\wedge 1)$ uses $(ϵ\wedge 1)\Delta =1_A$ and $\mu (\eta \wedge 1)=1_E$, leaving exactly evaluation. Its adjoint is $1_H$. The same calculation with $(1\wedge ϵ)\Delta$ proves the right unit. HenceIf $E$ is a commutative ring spectrum, cocommutativity of $\Delta$ proves that this function spectrum is commutative as well.