In the following, I'm following Markus Land, Introduction to Infinity-Categories (p. 81).
Let $\mathscr{C}$ be an $\infty$-category and $x,y,z \in \mathscr{C}$ be objects, then I want to understand the composition map $$\operatorname{map}_{\mathscr{C}}(x,y) \times \operatorname{map}_{\mathscr{C}}(y,z) \to \operatorname{map}_{\mathscr{C}}(x,z).$$ To do so, we have to define a composition map $$\operatorname{Fun}(\Delta^1, \mathscr{C}) \times_{\mathscr{C}} \operatorname{Fun}(\Delta^1, \mathscr{C}) \to \operatorname{Fun}(\Delta^1, \mathscr{C})$$ where for $\infty$-categories $\mathscr{D}, \mathscr{E}$ we define $\operatorname{Fun}(\mathscr{D}, \mathscr{E}) = \underline{\mathrm{Hom}}(\mathscr{D}, \mathscr{E})$ as the internal Hom of simplicial sets.
Question. How do we do this rigorously?
Land leaves this without a comment but there is something to do (at least for me at my current level). To define a map, we need to define compatible maps $$ \mathrm{Hom}_{\mathbf{sSet}}(\Delta^1 \times \Delta^k, \mathscr{C}) \times_{\mathscr{C}} \mathrm{Hom}_{\mathbf{sSet}}(\Delta^1 \times \Delta^k, \mathscr{C}) \to \mathrm{Hom}_{\mathbf{sSet}}(\Delta^1 \times \Delta^k, \mathscr{C})$$ for every $k \in \mathbb{N}$. If $k = 0$, then we can choose a composition since $\mathscr{C}$ is an $\infty$-category. (This is a non-canonical choice, but it is unique up to homotopy.) I suppose for higher $k$ we may start with the composition already inductively determined by the lower $k$ and try to fill the $(k+1)$-cells in a compatible way again.
But it seems messy to follow this approach. What's a transparent way of doing this?
Best Answer
Note that the domain of the composition map is isomorphic to $\operatorname{Fun}(\Lambda^2_1,\mathcal C)$. Therefore, you can define composition by choosing a homotopy inverse of the categorical equivalence $\operatorname{Fun}(\Delta^2,\mathcal C)\to \operatorname{Fun}(\Lambda^2_1,\mathcal C)$ and then compose with precomposition by $d^1\colon \Delta^1\to\Delta^2$.