# 3 Basics of Goodwillie calculus

The calculation of the derivative will use the explicit description of $P_ k F$ 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 $F$ by a $k$-excisive functor, we force $F$ 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 $\mathcal{X}: \mathbb {P}(S) \to \mathcal{C}$, where $|S| = k + 1$ and $\mathcal{X}(\emptyset ) = X$. We then replace $F(X)$ by $\lim _{\emptyset \neq T \subseteq S}F(\mathcal{X}(T))$ (which would equal $F(X)$ if $F$ were $k$-excisive). We repeat this procedure and hope we eventually end up with something $k$-excisive.

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

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

where the vertices are indexed by $T$.

We define

$T_ k F(X) = \lim _{\emptyset \neq T \subseteq S}F(C_ T(X)).$There is then a natural transformation $t_ F\colon F \to T_ k F$.

If $F\colon \mathcal{C} \to \mathcal{D}$ is any functor, define $P_ k F\colon \mathcal{C} \to \mathcal{D}$ as the sequential colimit

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

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

In the $1$-categorical case, to prove the universal property, we first need to know that $\theta _ F$ is an equivalence if $F$ is already $k$-excisive, which is clear in our case. This means if we have a map $\alpha : F \to G$ where $G$ is $k$-excisive, then we have a diagram

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

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

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

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 $P_ k(\theta _ F)$ is an equivalence, it suffices to show that $P_ k(t_ F: F \to T_ k F)$ 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 $P_ k F$ is in fact $k$-excisive.

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

The Cartesian $k$-cube $Y: \mathbb {P}(S) \to \mathcal{D}$ we seek admits a very simple description. Indeed, we simply take $F(\mathcal{X})$ 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 $F(\mathcal{X}) \to Y$ by the universal property.

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

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

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

We make the following observations:

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

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

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

The last fact gives us a map

$F(\mathcal{X}_ U(T)) \to F(C_ U(\mathcal{X}(T)))$natural in $U$ and $T$. These assemble to give maps

$F(\mathcal{X}(I)) \cong F(\mathcal{X}_{\emptyset }(I)) \to \lim _{\emptyset \not= T \subseteq S} F(\mathcal{X}_ T(I)) \to \lim _{\emptyset \not= T \subseteq S} F(C_ T(\mathcal{X}(I))) \cong T_ kF(\mathcal{X})(I)$It remains to show that the middle object is equal to $Y$. But it is not difficult to use the second fact to see that

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