# Kerodon

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### 8.4.7 Homotopy Idempotent Endomorphisms

Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $e: X \rightarrow X$ be an endomorphism in $\operatorname{\mathcal{C}}$. If $e$ is idempotent (in the sense of Definition 8.4.6.3), then the homotopy class $[e]$ is an idempotent endomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. One can ask if the converse is true: if the homotopy class $[e]$ is an idempotent endomorphism in $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$, does it follow that $e$ is an idempotent endomorphism of $\operatorname{\mathcal{C}}$? In this section, we will show that this question has a negative answer in general (Proposition 8.4.7.15), but a positive answer under some additional assumptions (Corollary 8.4.7.5). Let us begin by introducing some terminology.

Definition 8.4.7.1. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $e: X \rightarrow X$ be an endomorphism in $\operatorname{\mathcal{C}}$. We say that $e$ is homotopy idempotent if the homotopy class $[e]$ is an idempotent in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$, in the sense of Definition 8.4.2.1.

Remark 8.4.7.2. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $e: X \rightarrow X$ be an endomorphism in $\operatorname{\mathcal{C}}$. Then $e$ is homotopy idempotent if and only if there exists a $2$-simplex $\sigma$ of $\operatorname{\mathcal{C}}$ whose boundary is indicated in the diagram

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{e} & \\ X \ar [ur]^{e} \ar [rr]^{e} & & X. }$

Example 8.4.7.3. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $e: X \rightarrow X$ be an endomorphism in $\operatorname{\mathcal{C}}$. If $e$ is idempotent (that is, if it extends to a functor $\operatorname{N}_{\bullet }( \operatorname{Idem}) \rightarrow \operatorname{\mathcal{C}}$), then it is homotopy idempotent.

We now provide a partial converse to Example 8.4.7.3.

Proposition 8.4.7.4. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $e: X \rightarrow X$ be an endomorphism in $\operatorname{\mathcal{C}}$. The following conditions are equivalent:

$(1)$

The homotopy class $[e]$ is a split idempotent in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$.

$(2)$

The endomorphism $e$ is a split idempotent in $\operatorname{\mathcal{C}}$.

$(3)$

The endomorphism $e$ is homotopy idempotent and weakly split (Definition 8.4.6.14).

$(4)$

The endomorphism $e$ is homotopy idempotent and there exists a limit diagram

$\xymatrix@R =100pt@C=75pt{ & & \ar [dll] Y \ar [dl]^{i(-1)} \ar [d]^{ i(0)} \ar [dr]^{i(1)} \ar [drr] & & \\ \cdots \ar [r] & X \ar [r]^-{ e} & X \ar [r]^-{e} & X \ar [r]^-{e} & \cdots }$

in $\operatorname{\mathcal{C}}$, where the morphism $i(0)$ has a left homotopy inverse.

Proof. The equivalence $(1) \Leftrightarrow (2)$ is tautology, the implication $(2) \Rightarrow (3)$ follows from Proposition 8.4.6.17 (and Example 8.4.7.3), 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

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{e} & \\ X \ar [ur]^{e} \ar [rr]^{e} & & X. }$

Let $\operatorname{Spine}[\operatorname{\mathbf{Z}}]$ denote the $1$-dimensional simplicial set of Notation 8.4.4.12, 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 8.4.6.11. 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

$\xymatrix@R =50pt@C=50pt{ & & X \ar [d]^{r} & & \\ & & Y \ar [dl]_{ i(-1) } \ar [d]^{ i(0)} \ar [dr]^{ i(1) } & & \\ \cdots \ar [r] & X \ar [r]^-{ e} & X \ar [r]^-{e} & X \ar [r]^-{e} & \cdots }$

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

\begin{eqnarray*} [i(0)] \circ ([r] \circ [i(0)]) & = & ( [i(0)] \circ [r] ) \circ [i(0)] \\ & = & [e] \circ [i(0)] \\ & = & [e] \circ ( [e] \circ [i(-1) ] ) \\ & = & ( [e] \circ [e] ) \circ [i(-1) ] \\ & = & [e] \circ [i(-1) ] \\ & = & [i(0)]. \end{eqnarray*}
$\square$

Corollary 8.4.7.5. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category which admits sequential limits and colimits, and let $e: X \rightarrow X$ be an endomorphism in $\operatorname{\mathcal{C}}$. Then $e$ is idempotent if and only if it is homotopy idempotent and the composite map

$\varprojlim ( \cdots \rightarrow X \xrightarrow {e} X \rightarrow \cdots ) \rightarrow X \rightarrow \varinjlim ( \cdots \rightarrow X \xrightarrow {e} X \rightarrow \cdots )$

is an isomorphism (see Definition 8.4.6.14).

Proof. This is a special case of Proposition 8.4.7.4, together with the observation that every idempotent in $\operatorname{\mathcal{C}}$ is split (Proposition 8.4.6.12). $\square$

Let us now present a sample application of Proposition 8.4.7.4.

Corollary 8.4.7.6. Let $X$ be a connected Kan complex and let $x \in X$ be a vertex. Then every homotopy idempotent endomorphism $e: (X,x) \rightarrow (X,x)$ in the $\infty$-category $\operatorname{\mathcal{S}}_{\ast }$ is (split) idempotent.

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 5.6.3.8). Then the diagram $T_{e}: \operatorname{Spine}[\operatorname{\mathbf{Z}}] \rightarrow \operatorname{\mathcal{S}}_{\ast }$ of Notation 8.4.6.11 lifts to a functor of ordinary categories $\mathscr {F}: (\operatorname{\mathbf{Z}}, \leq ) \rightarrow \operatorname{Kan}_{\ast }$, which we display as

$\cdots \rightarrow (X,x) \xrightarrow {e} (X,x) \xrightarrow {e} (X,x) \xrightarrow {e} (X,x) \xrightarrow {e} (X,x) \rightarrow \cdots$

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 8.4.4.14), 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 7.1.3.20, 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 $\infty$-category $\operatorname{N}_{\bullet }(\operatorname{\mathbf{Z}})$ is filtered 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 7.5.9.3). Since the spine $\operatorname{Spine}[\operatorname{\mathbf{Z}}]$ is weakly contractible (Remark 8.4.4.15), 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

$\Delta ^1 \simeq \{ - \infty \} \star \{ \infty \} \hookrightarrow \{ - \infty \} \star \operatorname{Spine}[\operatorname{\mathbf{Z}}] \star \{ \infty \} \xrightarrow {S} \operatorname{\mathcal{S}}_{\ast }$

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

$\theta : \mathscr {G}^{\pm }(-\infty ) = \varprojlim ( \mathscr {G} ) \rightarrow \varinjlim ( \mathscr {G} ) = \mathscr {G}^{\pm }( \infty )$

is a homotopy equivalence of (pointed) Kan complexes (Corollary 8.4.7.6).

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

$\cdots \rightarrow \pi _{d}( \mathscr {G}(-1), y_{-1} ) \rightarrow \pi _{d}( \mathscr {G}(0), y_0 ) \rightarrow \pi _{d}( \mathscr {G}(1), y_1 ) \rightarrow \cdots$

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

$\cdots \rightarrow \pi _{d}(X,x) \xrightarrow {f_ d} \pi _{d}(X,x) \xrightarrow {f_ d} \pi _{d}(X,x) \rightarrow \cdots$

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

$\pi _{d}( \varprojlim _{n} \mathscr {G}(n), y_{\infty } ) \xrightarrow {u} \varprojlim _{n} \pi _{d}( \mathscr {G}(n), y_ n) ) \xrightarrow {v} \varinjlim _{n} \pi _{d}(\mathscr {G}(n), y_ n ) \xrightarrow {w} \pi _{d}( \mathscr {G}(\infty ), y_{\infty } ),$

where the map $w$ is also bijective (Remark 3.2.2.15). 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

$\operatorname{ {\varprojlim }^{1} }( \cdots \rightarrow \pi _{d+1}( \mathscr {G}(-2), y_{-2} ) \rightarrow \pi _{d+1}( \mathscr {G}(-1), y_{-1} ) \rightarrow \pi _{d+1}( \mathscr {G}(0), y_0 ))$

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

$\cdots \rightarrow \pi _{d+1}(X,x) \xrightarrow { f_{d+1} } \pi _{d+1}(X,x) \xrightarrow { f_{d+1} } \pi _{d+1}( X,x )$

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

Corollary 8.4.7.7. Let $X$ be a connected Kan complex and let $e: X \rightarrow X$ be a homotopy idempotent endomorphism in the $\infty$-category $\operatorname{\mathcal{S}}$. Then $e$ is idempotent if and only if it can be lifted to a homotopy endomorphism endomorphism $\widetilde{e}: (X,x) \rightarrow (X,x)$ in the $\infty$-category $\operatorname{\mathcal{S}}_{\ast }$.

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 8.4.7.6), 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 7.4.5.6), the idempotent $F$ splits. Consequently, there is a retraction diagram

8.30
$$\begin{gathered}\label{equation:non-coherent-idempotent} \xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{r} & \\ Y \ar [ur]^{i} \ar [rr]^{ \operatorname{id}_{Y} } & & Y } \end{gathered}$$

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 5.6.3.4). Since the forgetful functor $\operatorname{\mathcal{S}}_{\ast } \rightarrow \operatorname{\mathcal{S}}$ is a left fibration, we can lift (8.30) to a retraction diagram

$\xymatrix@R =50pt@C=50pt{ & (X,x) \ar [dr]^{ \widetilde{r} } & \\ (Y,y) \ar [ur]^{ \widetilde{i} } \ar [rr]^{ \operatorname{id}} & & (Y,y) }$

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 $\widetilde{i}$. $\square$

Exercise 8.4.7.8. Let $(X,x)$ be a pointed Kan complex and let $e: X \rightarrow X$ be a morphism from $X$ to itself. Show that:

• If $X$ is connected, then $e$ can be lifted to a morphism $\widetilde{e}: (X,x) \rightarrow (X,x)$ in the $\infty$-category $\operatorname{\mathcal{S}}_{\ast }$.

• If $X$ is simply connected, then $e$ is homotopy idempotent (in the $\infty$-category $\operatorname{\mathcal{S}}$) if and only if $\widetilde{e}$ is homotopy idempotent (in the $\infty$-category $\operatorname{\mathcal{S}}_{\ast }$).

In particular, if $X$ is simply connected, then every homotopy idempotent $e: X \rightarrow X$ is (split) idempotent (Corollary 8.4.7.7).

In the situation of Exercise 8.4.7.8, the simple connectivity assumption on $X$ cannot be omitted. That is, not every homotopy idempotent in the $\infty$-category $\operatorname{\mathcal{S}}$ is split. We present a counterexample, due originally to Freyd and Heller (see ).

Definition 8.4.7.9 (Dyadic Homeomorphisms). Recall that a dyadic rational number is a real number of the form $\frac{a}{2^ n}$, where $a$ and $n$ are integers. Let $s, t \geq 0$ be dyadic rational numbers. We say that a homeomorphism $f: [0,s] \xrightarrow {\sim } [0,t]$ is dyadic if it satisfies the following conditions:

• The function $f$ is piecewise linear; in particular, it is differentiable away from finitely many points of the closed interval $[0,s]$.

• If $x \in [0,s]$ is a point where $f$ is not differentiable, then $x$ is a dyadic rational number.

• For every point $x \in [0,s]$ where $f$ is differentiable, the derivative $f'(x)$ is equal to $2^{n}$ for some integer $n$.

Note that the third condition implies that the homeomorphism $f$ is strictly increasing, so that $f(0) = 0$ and $f(s) = t$.

Exercise 8.4.7.11 (Composition of Dyadic Homeomorphisms). Let $s,t,u \geq 0$ be dyadic rational numbers and let $f: [0,s] \xrightarrow {\sim } [0,t]$ and $g: [0,t] \xrightarrow {\sim } [0,u]$ be dyadic homeomorphisms. Show that the composition $(g \circ f): [0,s] \xrightarrow {\sim } [0,u]$ is also a dyadic homeomorphism.

Definition 8.4.7.12 (The Thompson Group). Let $\operatorname{Aut}_{ \mathrm{Dy} }( [0,1] )$ denote the collection of all dyadic homeomorphisms from the unit interval $[0,1]$ to itself. It follows from Exercises 8.4.7.10 and 8.4.7.11 that $\operatorname{Aut}_{ \mathrm{Dy} }( [0,1] )$ has the structure of a group (where the group law is given by composition of homeomorphisms). We will refer to $\operatorname{Aut}_{ \mathrm{Dy} }( [0,1] )$ as the Thompson group.

Construction 8.4.7.13 (Speeding Up). Let $f: [0,1] \rightarrow [0,1]$ be an orientation-preserving homeomorphism. We define $\alpha (f): [0,1] \rightarrow [0,1]$ by the formula

$\alpha (f)(x) = \begin{cases} f(2x) / 2 & \text{ if 0 \leq x \leq 1/2} \\ x & \text{ if 1/2 \leq x \leq 1. } \end{cases}$

Then $\alpha (f)$ is also an orientation-preserving homeomorphism of $[0,1]$ with itself. Moreover, if $f$ is dyadic, then $\alpha (f)$ is also dyadic. It follows that the construction $f \mapsto \alpha (f)$ determines a group homomorphism $\alpha$ from the Thompson group $\operatorname{Aut}_{ \mathrm{Dy} }([0,1])$ to itself.

Proposition 8.4.7.14. Let $\operatorname{Aut}_{ \mathrm{Dy} }([0,1])$ be the Thompson group of Definition 8.4.7.12 and let $X = B_{\bullet } \operatorname{Aut}_{ \mathrm{Dy} }([0,1])$ denote its classifying simplicial set (Example 1.2.4.3). Then the homomorphism $\alpha$ of Construction 8.4.7.13 induces a homotopy idempotent endomorphism $e: X \rightarrow X$ in the $\infty$-category $\operatorname{\mathcal{S}}$.

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

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{e} & \\ X \ar [ur]^{e} \ar [rr]^{e } & & X }$

commutes up to homotopy. By virtue of Proposition 1.4.3.3, 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 $0 \leq x \leq 1/4$. $\square$

We now show that the homotopy idempotent of Proposition 8.4.7.14 cannot be refined to an idempotent in the $\infty$-category $\operatorname{\mathcal{S}}$.

Proposition 8.4.7.15. Let $X = B_{\bullet } \operatorname{Aut}_{ \mathrm{Dy} }([0,1])$. Then homotopy idempotent endomorphism $e: X \rightarrow X$ of Proposition 8.4.7.14 is not idempotent.

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 8.4.7.7). 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 8.4.7.13. 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). $\square$