Goodwillie filtration and factorization homology — Basics of Goodwillie calculus

3 Basics of Goodwillie calculus

The calculation of the derivative will use the explicit description of PkFP_ k F in the proof of its existence, which is the focus of this section.

We keep the assumptions of the previous section. To construct the universal approximation of FF by a kk-excisive functor, we force FF to send certain strongly coCartesian diagrams to Cartesian diagrams, and it will turn out that this is enough.

More specifically, we find some strongly coCarteisan diagrams X:P(S)C\mathcal{X}: \mathbb {P}(S) \to \mathcal{C}, where S=k+1|S| = k + 1 and X()=X\mathcal{X}(\emptyset ) = X. We then replace F(X)F(X) by limTSF(X(T))\lim _{\emptyset \neq T \subseteq S}F(\mathcal{X}(T)) (which would equal F(X)F(X) if FF were kk-excisive). We repeat this procedure and hope we eventually end up with something kk-excisive.

To define such an X\mathcal{X}, we need to specify XP=1(S)\mathcal{X}|_{\mathbb {P}_{=1}(S)}, since we know X()=X\mathcal{X}(\emptyset ) = X and everything is the left Kan extension of this. Since XX is the only thing we know, the only thing we can do is to set X({s})=\mathcal{X}(\{ s\} ) = * (placing XX there does no good).

We define C()(X):P(S)CC_{(-)}(X): \mathbb {P}(S) \to \mathcal{C} to be the unique strongly coCartesian diagram such that C(X)=XC_{\emptyset }(X) = X and C{s}=C_{\{ s\} } = *. Concretely, CT(X)C_ T(X) is the colimit of the diagram

\begin{useimager} 
  \[
    \begin{tikzcd}
      & & X \ar[dll] \ar[dl] \ar[d] \ar[dr] \ar[drr]\\
      * & * & \cdots & *& *
    \end{tikzcd}
  \]
\end{useimager}

where the vertices are indexed by TT.

We define

TkF(X)=limTSF(CT(X)). T_ k F(X) = \lim _{\emptyset \neq T \subseteq S}F(C_ T(X)).

There is then a natural transformation tF ⁣:FTkFt_ F\colon F \to T_ k F.

Theorem 19

If F ⁣:CDF\colon \mathcal{C} \to \mathcal{D} is any functor, define PkF ⁣:CDP_ k F\colon \mathcal{C} \to \mathcal{D} as the sequential colimit

\begin{useimager} 
    \[
      \begin{tikzcd}
        F \ar[r, "t_F"] & T_k F \ar[r, "t_{T_k F}"] & T_k T_k F \ar[r] & \cdots
      \end{tikzcd}
    \]
  \end{useimager}

Then PkFP_ k F is kk-excisive and the natural map θF:FPkF\theta _ F: F \to P_ k F is the universal natural transformation to such a functor.

Proof

We first figure out what we have to prove. Of course, we have to prove that PkFP_ k F is kk-excisive. It turns out this is the only non-trivial thing to prove.

In the 11-categorical case, to prove the universal property, we first need to know that θF\theta _ F is an equivalence if FF is already kk-excisive, which is clear in our case. This means if we have a map α:FG\alpha : F \to G where GG is kk-excisive, then we have a diagram

\begin{useimager} 
    \[
      \begin{tikzcd}
        F \ar[r, "\alpha"] \ar[d, "\theta_F"] & G \ar[d, "\theta_G"]\\
        P_k F \ar[r, "P_k \alpha"] & P_k G
      \end{tikzcd}
    \]
  \end{useimager}

Since θG\theta _ G is an equivalence, PkαP_ k\alpha lifts to a map to GG that makes the diagram commute.

To show that the extension to PkFP_ k F is unique, suppose we have two extensions α~,α~\tilde{\alpha }, \tilde{\alpha }'

\begin{useimager} 
    \[
      \begin{tikzcd}
        F \ar[r, "\alpha"] \ar[d, "\theta_F"] & G\\
        P_k F \ar[ur, dashed]
      \end{tikzcd}
    \]
  \end{useimager}

Applying PkP_ k to the whole diagram, we know that Pk(α~)Pk(θF)=Pk(α~)Pk(θF)P_ k(\tilde{\alpha }) \circ P_ k(\theta _ F) = P_ k(\tilde{\alpha }') \circ P_ k(\theta _ F). If Pk(θF)P_ k(\theta _ F) were an equivalence, then this implies Pk(α~)=Pk(α~)P_ k(\tilde{\alpha }) = P_ k(\tilde{\alpha }'). Since PkP_ k essentially acts as the identity on PkFP_ k F and GG, this implies α~=α~\tilde{\alpha } = \tilde{\alpha }'.

In the \infty -categorical case, HTT 5.2.7.4 implies these two conditions are also sufficient.

As we said, the first condition is immediate from construction, and to prove that Pk(θF)P_ k(\theta _ F) is an equivalence, it suffices to show that Pk(tF:FTkF)P_ k(t_ F: F \to T_ k F) is an equivalence. But this is clear, since we are just shifting the sequential colimit.

So all we have to do is to show that PkFP_ k F is in fact kk-excisive.

Claim

Let X ⁣:P(S)C\mathcal{X}\colon \mathbb {P}(S) \to \mathcal{C} be a strongly coCartesian kk-cube. Then the canonical map θF ⁣:F(X)(TkF)(X)\theta _ F\colon F(\mathcal{X}) \to (T_ kF)(\mathcal{X}) factors through a Cartesian kk-cube in D\mathcal{D}.

If this were true, then PkF(X)P_ k F(\mathcal{X}) would be the sequential colimit of Cartesian cubes, hence Cartesian (since we assumed finite limits commute with sequential colimits).

The Cartesian kk-cube Y:P(S)DY: \mathbb {P}(S) \to \mathcal{D} we seek admits a very simple description. Indeed, we simply take F(X)F(\mathcal{X}) and replace the \emptyset vertex with the pullback of the rest of the diagram. This is then by construction a Cartesian cube, and there is a canonical map F(X)YF(\mathcal{X}) \to Y by the universal property.

To show that θF\theta _ F factors through this map, we need to use a funny description of YY, and this description uses the fact that X\mathcal{X} is strongly coCartesian.

Fix a TST \subseteq S. We define XT:P(S)C\mathcal{X}_ T: \mathbb {P}(S) \to \mathcal{C} by setting XT(I)\mathcal{X}_ T(I) to be the pushout of the diagram

\begin{useimager} 
    \[
      \begin{tikzcd}[column sep=small]
        & & \mathcal{X}(I) \ar[dll] \ar[dl] \ar[d] \ar[dr] \ar[drr]\\
        \mathcal{X}(I \cup \{s_1\}) & \mathcal{X}(I \cup \{s_2\}) & \cdots & \mathcal{X}(I \cup \{s_{k - 1}\}) & \mathcal{X}(I \cup \{s_k\})
      \end{tikzcd}
    \]
  \end{useimager}

where T={s1,,sk}T = \{ s_1, \ldots , s_ k\} .

We make the following observations:

  • X(I)=X(I)\mathcal{X}_{\emptyset }(I) = \mathcal{X}(I).

  • If X\mathcal{X} is strongly coCartesian, then essentially by definition, XT(I)=X(IT)\mathcal{X}_ T(I) = \mathcal{X}(I \cup T).

  • If we replace the bottom vertices with *, then this gives CT(X(I))C_ T(\mathcal{X}(I)). So there is a canonical map XT(I)CT(X(I))\mathcal{X}_ T(I) \to C_ T(\mathcal{X}(I)).

The last fact gives us a map

F(XU(T))F(CU(X(T))) F(\mathcal{X}_ U(T)) \to F(C_ U(\mathcal{X}(T)))

natural in UU and TT. These assemble to give maps

F(X(I))F(X(I))limTSF(XT(I))limTSF(CT(X(I)))TkF(X)(I) F(\mathcal{X}(I)) \cong F(\mathcal{X}_{\emptyset }(I)) \to \lim _{\emptyset \not= T \subseteq S} F(\mathcal{X}_ T(I)) \to \lim _{\emptyset \not= T \subseteq S} F(C_ T(\mathcal{X}(I))) \cong T_ kF(\mathcal{X})(I)

It remains to show that the middle object is equal to YY. But it is not difficult to use the second fact to see that

limTSF(XT(I))={limTSF(X(T))I=F(X(I))I. \lim _{\emptyset \neq T \subseteq S}F(\mathcal{X}_ T(I)) = \begin{cases} \lim _{\emptyset \neq T \subseteq S}F(\mathcal{X}(T)) & I = \emptyset \\ F(\mathcal{X}(I)) & I \neq \emptyset \end{cases}.\qedhere