Kerodon

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Remark 7.6.5.5. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category, and suppose we are given a tower $X: \operatorname{N}_{\bullet }( \operatorname{\mathbf{Z}}_{\geq 0}^{\operatorname{op}} ) \rightarrow \operatorname{\mathcal{C}}$, which we depict as a diagram

\[ \cdots \rightarrow X(4) \xrightarrow { f_3 } X(3) \xrightarrow { f_2 } X(2) \xrightarrow { f_1} X(1) \xrightarrow { f_0 } X(0), \]

having a limit $\varprojlim (X)$. Then, for every object $Y \in \operatorname{\mathcal{C}}$, the map of sets

\[ \theta : \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( Y, \varprojlim (X) ) \rightarrow \varprojlim ( \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( Y, X(n) ) ) \]

is surjective. To prove this, suppose we are given a collection of morphisms $g_ n: Y \rightarrow X(n)$ satisfying $[f_{n}] \circ [g_{n+1} ] = [ g_ n ]$ in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. Then, for each $n \geq 0$, we can choose a $2$-simplex $\sigma _ n$ in $\operatorname{\mathcal{C}}$ as indicated in the diagram

\[ \xymatrix@C =50pt@R=50pt{ & X(n+1) \ar [dr]^-{ f_ n } & \\ Y \ar [ur]^{ g_{n+1} } \ar [rr]^-{ g_ n } & & X(n). } \]

Let $X_0$ denote the restriction of $X$ to the spine $\operatorname{Spine}[ \operatorname{\mathbf{Z}}_{\geq 0}^{\operatorname{op}} ] \subset \operatorname{N}_{\bullet }( \operatorname{\mathbf{Z}}_{\geq 0}^{\operatorname{op}} )$. Then the collection of $2$-simplices $\{ \sigma _ n \} _{n \geq 0}$ determines an extension of $X_0$ to a diagram $\overline{X}_0: \operatorname{Spine}[ \operatorname{\mathbf{Z}}_{\geq 0}^{\operatorname{op}} ]^{\triangleleft } \rightarrow \operatorname{\mathcal{C}}$ carrying the cone point to the object $Y$. The isomorphism class of this extension can be identified with a morphism $[g]: Y \rightarrow \varprojlim (X_0) \simeq \varprojlim (X)$ in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$, which is a preimage of the sequence $\{ [g_ n] \} _{n \geq 0}$ under the function $\theta $.