Kerodon

Proof. The equivalence $(1) \Leftrightarrow (2)$ is tautology, the implication $(2) \Rightarrow (3)$ follows from Proposition 8.5.6.17 (and Example 8.5.7.3), and the implication $(3) \Rightarrow (4)$ is immediate. We will complete the proof by showing that $(4)$ implies $(1)$. Assume that $e$ is homotopy idempotent, so that there exists a $2$-simplex $\sigma $ of $\operatorname{\mathcal{C}}$ whose boundary is indicated in the diagram

Let $\operatorname{Spine}[\operatorname{\mathbf{Z}}]$ denote the $1$-dimensional simplicial set of Notation 8.5.4.12, and let $T_{\sigma }: \operatorname{Spine}[\operatorname{\mathbf{Z}}] \rightarrow \operatorname{\mathcal{C}}_{X/}$ be the morphism of simplicial sets which carries each vertex of $\operatorname{Spine}[\operatorname{\mathbf{Z}}]$ to $e$ (regarded as an object of the coslice $\infty $-category $\operatorname{\mathcal{C}}_{X/}$) and each nondegenerate edge of $\operatorname{Spine}[\operatorname{\mathbf{Z}}]$ to $\sigma $ (regarded as a morphism of the coslice $\infty $-category $\operatorname{\mathcal{C}}_{X/}$). We will identify $T_{\sigma }$ with a diagram $\overline{T}_{e}: \operatorname{Spine}[\operatorname{\mathbf{Z}}]^{\triangleleft } \rightarrow \operatorname{\mathcal{C}}$, where the restriction $\overline{T}_{e}|_{ \operatorname{Spine}[\operatorname{\mathbf{Z}}] }$ is the diagram $T_{e}: \operatorname{Spine}[\operatorname{\mathbf{Z}}] \rightarrow \operatorname{\mathcal{C}}$ of Notation 8.5.6.11. Let us further identify $\overline{T}_{e}$ with an object $\overline{X}$ of the slice $\infty $-category $\operatorname{\mathcal{C}}_{ / T_{e} }$, lying over the object $X \in \operatorname{\mathcal{C}}$. By virtue of assumption $(4)$, the $\infty $-category $\operatorname{\mathcal{C}}_{ / T_{e} }$ has a final object $\overline{Y}$, lying over a some object $Y \in \operatorname{\mathcal{C}}$. We can therefore choose a morphism $\overline{r}: \overline{X} \rightarrow \overline{Y}$ in the $\infty $-category $\operatorname{\mathcal{C}}_{ / T_{e} }$, having some image $r: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$. The morphism $\overline{r}$ can be identified with a diagram $\Delta ^1 \star \operatorname{Spine}[\operatorname{\mathbf{Z}}] \rightarrow \operatorname{\mathcal{C}}$, which we display informally as

By construction, the restriction of this diagram to the middle column witnesses the equality $[i(0)] \circ [r] = [e]$ in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. To show that the homotopy class $[e]$ is a split idempotent, it will suffice to show that $[r] \circ [i(0)]$ is the identity morphism $[ \operatorname{id}_{Y} ]$ in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. Since the morphism $[i(0)]$ admits a left inverse, it will suffice to show that equality holds after postcomposition with $[i(0)]$. This follows from the calculation

Proof. This is a special case of Proposition 8.5.7.4, together with the observation that every idempotent in $\operatorname{\mathcal{C}}$ is split (Proposition 8.5.6.12). $\square$

Proof. Without loss of generality, we may assume that the morphism $e$ is obtained from a morphism $(X,x) \rightarrow (X,x)$ in the ordinary category of pointed Kan complexes (see Proposition 5.5.3.8). Then the diagram $T_{e}: \operatorname{Spine}[\operatorname{\mathbf{Z}}] \rightarrow \operatorname{\mathcal{S}}_{\ast }$ of Notation 8.5.6.11 lifts to a functor of ordinary categories $\mathscr {F}: (\operatorname{\mathbf{Z}}, \leq ) \rightarrow \operatorname{Kan}_{\ast }$, which we display as

Let us abuse notation by identifying $\mathscr {F}$ with its image in the category of Kan complexes $\operatorname{Kan}$. Applying Variant 7.5.3.6, we can choose a levelwise homotopy equivalence $\alpha : \mathscr {F} \rightarrow \mathscr {G}$, where $\mathscr {G}: (\operatorname{\mathbf{Z}}, \leq ) \rightarrow \operatorname{Kan}$ is an isofibrant diagram of Kan complexes. Note that we can also regard $\mathscr {G}$ as a diagram of pointed Kan complexes, by equipping each $\mathscr {G}(n)$ with the base point $y_{n} = \alpha (n)(x)$. Let us extend $\mathscr {G}$ to a functor $\mathscr {G}^{\pm }: ( \operatorname{\mathbf{Z}}\cup \{ -\infty , \infty \} , \leq ) \rightarrow \operatorname{Kan}_{\ast }$ by setting $\mathscr {G}^{\pm }(-\infty ) = \varprojlim ( \mathscr {G} )$ and $\mathscr {G}^{\pm }(\infty ) = \varinjlim ( \mathscr {G} )$, where the limit and colimit are formed in the category of (pointed) simplicial sets; we denote the base points of $\mathscr {G}^{\pm }(-\infty )$ and $\mathscr {G}^{\pm }(\infty )$ by $y_{-\infty }$ and $y_{\infty }$, respectively. Passing to nerves, the functor $\mathscr {G}^{\pm }$ determines a diagram $S: \{ - \infty \} \star \operatorname{Spine}[\operatorname{\mathbf{Z}}] \star \{ \infty \} \rightarrow \operatorname{\mathcal{S}}_{\ast }$.

Let $U: \operatorname{\mathcal{S}}_{\ast } \rightarrow \operatorname{\mathcal{S}}$ be the forgetful functor (given on objects by $U(X,x) = X$). Since the diagram $\mathscr {G}$ is isofibrant and the inclusion $\operatorname{Spine}[\operatorname{\mathbf{Z}}] \hookrightarrow \operatorname{N}_{\bullet }(\operatorname{\mathbf{Z}})$ is left cofinal (Remark 8.5.4.14), the restriction $(U \circ S)|_{ \{ - \infty \} \star \operatorname{Spine}[\operatorname{\mathbf{Z}}] }$ is a limit diagram in the $\infty $-category $\operatorname{\mathcal{S}}$ (Corollary 7.5.4.7). Applying Corollary 7.1.4.21, we see that $S|_{ \{ - \infty \} \star \operatorname{Spine}[\operatorname{\mathbf{Z}}] }$ is a limit diagram in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$. Since the partially ordered set $(\operatorname{\mathbf{Z}}, \geq )$ is directed and the inclusion $\operatorname{Spine}[\operatorname{\mathbf{Z}}] \hookrightarrow \operatorname{N}_{\bullet }(\operatorname{\mathbf{Z}})$ is right cofinal, the restriction $(U \circ S)|_{ \operatorname{Spine}[\operatorname{\mathbf{Z}}] \star \{ \infty \} }$ is a colimit diagram in the $\infty $-category $\operatorname{\mathcal{S}}$ (Corollary 9.1.7.3). Since the spine $\operatorname{Spine}[\operatorname{\mathbf{Z}}]$ is weakly contractible (Remark 8.5.4.15), it follows that $S|_{ \operatorname{Spine}[\operatorname{\mathbf{Z}}] \star \{ \infty \} }$ is a colimit diagram in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$. Moreover, the natural transformation $\alpha $ induces an isomorphism $T_{e} \rightarrow S|_{ \operatorname{Spine}[\operatorname{\mathbf{Z}}] }$ in the $\infty $-category $\operatorname{Fun}( \operatorname{Spine}[\operatorname{\mathbf{Z}}], \operatorname{\mathcal{S}}_{\ast } )$. It follows that the morphism $e$ is idempotent if and only if the composition

is an isomorphism in $\operatorname{\mathcal{S}}_{\ast }$: that is, if and only if the map of Kan complexes

is a homotopy equivalence of (pointed) Kan complexes (Corollary 8.5.7.6).

Since each $\mathscr {G}(n)$ is a connected Kan complex, it follows that the colimit $\varinjlim ( \mathscr {G} )$ is also connected. By virtue of Theorem 3.2.7.1, it will suffice to show that, for every integer $d \geq 0$, $\theta $ induces a bijection $\pi _{d}( \varprojlim (\mathscr {G}), y_{-\infty } ) \rightarrow \pi _{d}( \varinjlim ( \mathscr {G}), y_{\infty } )$ (note that, in the case $d = 0$, this guarantees that the Kan complex $\varprojlim (\mathscr {G} )$ is also connected, so that a similar conclusion holds for any choice of base point). Let $\overleftarrow {G}$ denote the diagram of sets

Note that $\alpha $ determines an isomorphism of $\overleftarrow {G}$ with the diagram

where each of the transition maps is induced by $e$. Since $e$ is homotopic to $e \circ e$ (in the homotopy category of pointed Kan complexes), it follows that $f_ d = f_ d \circ f_ d$, so that the tautological map $v: \varprojlim ( \overleftarrow {G} ) \rightarrow \varinjlim ( \overleftarrow {G} )$ is a bijection. Unwinding the definition, we see that $\pi _{d}(\theta )$ factors as a composition

where the map $w$ is also bijective (Remark 3.2.2.16). It will therefore suffice to show that the map $u$ is bijective. By virtue of the Milnor exact sequence (Proposition ), this is equivalent to the assertion that the set

has a single element. This is a special case of Proposition , since the inverse system of groups

is Mittag-Leffler (since $f_{d+1}$ is idempotent, its image coincides with the image of $f_{d+1}^{n}$ for every integer $n > 0$). $\square$

Proof. Let $\widetilde{e}: (X,x) \rightarrow (X,x)$ be a lift of $e$ to a morphism in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$. If $\widetilde{e}$ is homotopy idempotent, then it is idempotent (Corollary 8.5.7.6), so that $e$ is also idempotent. For the converse, suppose that $e$ is idempotent: that is, it can be extended to a functor $F: \operatorname{N}_{\bullet }( \operatorname{Idem}) \rightarrow \operatorname{\mathcal{C}}$. Since the $\infty $-category $\operatorname{\mathcal{S}}$ admits small colimits (Corollary 7.4.3.3), the idempotent $F$ splits. Consequently, there is a retraction diagram

in the $\infty $-category of spaces $\operatorname{\mathcal{S}}$, where $e$ is homotopic to the composition $(i \circ r): X \rightarrow X$. Fix vertices $x \in X$ and $y \in Y$. Since $X$ is connected, we can lift $i$ to a morphism $\widetilde{i}: (Y,y) \rightarrow (X,x)$ in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$ (see Example 5.5.3.4). Since the forgetful functor $\operatorname{\mathcal{S}}_{\ast } \rightarrow \operatorname{\mathcal{S}}$ is a left fibration, we can lift (8.67) to a retraction diagram

in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$. It follows that $e$ can be lifted to a (split) homotopy idempotent in $\operatorname{\mathcal{S}}_{\ast }$, given by any composition of $\widetilde{r}$ with $\widetilde{i}$. $\square$

Proof. We wish to show that the diagram of Kan complexes

commutes up to homotopy. By virtue of Proposition 1.5.3.3, this is equivalent to the assertion that the homomorphisms $\alpha , \alpha ^2: \operatorname{Aut}_{ \mathrm{Dy} }([0,1]) \rightarrow \operatorname{Aut}_{ \mathrm{Dy} }([0,1])$ are conjugate: that is, there exists an element $g \in \operatorname{Aut}_{ \mathrm{Dy} }([0,1])$ satisfying the identity $\alpha (f) \circ g = g \circ \alpha ^2(f)$ for every element $f \in \operatorname{Aut}_{ \mathrm{Dy}}( [0,1] )$. Concretely, we can take $g$ to be any dyadic homeomorphism satisfying the identity $g(x) = 2x$ for $0 \leq x \leq 1/4$. $\square$

Proof. Let $x$ denote the unique vertex of $X$. Suppose, for a contradiction, that $e$ is idempotent. Then we can lift $e$ to a homotopy idempotent morphism $\widetilde{e}: (X,x) \rightarrow (X,x)$ in the $\infty $-category $\operatorname{\mathcal{S}}_{\ast }$ (Corollary 8.5.7.7). Passing to fundamental groups, we obtain an idempotent homomorphism $\beta $ from the Thompson group $\operatorname{Aut}(_{ \mathrm{Dy} }( [0,1] ) = \pi _{1}(X,x)$ to itself. Since the forgetful functor $\operatorname{\mathcal{S}}_{\ast } \rightarrow \operatorname{\mathcal{S}}$ carries $\widetilde{e}$ to $e$, $\beta $ is conjugate to the homomorphism $\alpha $ of Construction 8.5.7.13. Since $\alpha $ is a monomorphism, it follows that $\beta $ is also a monomorphism. The equation $\beta ^2 = \beta $ then implies that $\beta $ is the identity map. This is a contradiction, since $\beta $ is conjugate to the homomorphism $\alpha $ (which is not the identity morphism). $\square$