Kerodon

We need to show that the canonical map $\mathrm h(S_\bullet \times T_\bullet)\to \mathrm h S_\bullet \times \mathrm h T_\bullet$ is an isomorphism. And then you prove, using Yoneda, that a specific map $\mathrm h(S_\bullet \times T_\bullet)\to \mathrm h S_\bullet \times \mathrm h T_\bullet$ is an isomorphism. But why does that map coincide with the canonical map? That is, how do we see that "this map is given by the composition", in your words? A priori it could be another one.

Since this also took me some time to unwind, let me put the missing part of the argument here for the sake of future readers, maybe this could be added to the text.

Given a map $hS_{\bullet}\times hT_{\bullet}\rightarrow C$, we need to show that in the penultimate step we get the composition $S_{\bullet}\times T_{\bullet} \rightarrow N_{\bullet}h(S \times T) \rightarrow N_{\bullet}(hS\times hT) \rightarrow N_{\bullet}C$. The penultimate step in the composition gives us the composition $S_{\bullet}\times T_{\bullet} \rightarrow N_{\bullet}hS\times T_{\bullet} \rightarrow N_{\bullet}(Fun(hT,C))\times T_{\bullet} \rightarrow Fun(T_{\bullet},N_{\bullet}C))\times T_{\bullet} \rightarrow N_{\bullet}C$. Now the key point is that by the construction of the isomorphism in corollary 1.4.3.5, we see that the composition $N_{\bullet}(Fun(hT,C))\times T_{\bullet} \rightarrow Fun(T_{\bullet},N_{\bullet}C))\times T_{\bullet} \rightarrow N_{\bullet}C$ is actually just the composition $N_{\bullet}(Fun(hT,C))\times T_{\bullet} \rightarrow N_{\bullet}(Fun(hT,C)\times hT) \rightarrow N_{\bullet}C$ where the first map is given by identity times the unit and the second is just applying the nerve functor to the ordinary evaluation map. Thus we may transform our earlier composition into $S_{\bullet}\times T_{\bullet} \rightarrow N_{\bullet}(hS\times hT) \rightarrow N_{\bullet}(Fun(hT,C)\times hT) \rightarrow N_{\bullet}C$ (some identifications omitted) and now the result is clear by classical category theory,