# Extension of a functor by colimits: Cisinski - Higher Categories and Homotopical Algebra - Remark 1.1.11

First, I state premilinary results.

For a presheaf $$X\colon A^{op}\to\mathsf{Set}$$, it's category of elements, denoted by $$\int X$$, has pairs $$(a,s)$$ where $$a \in A$$ and $$s \in X(a)$$ as objects and $$f\colon a\to b$$ such that $$X(f)(t) = s$$ as morphisms $$(a,s)\to (b,t)$$.

Cisinski, Proposition 1.1.8 For each presheaf $$X$$ over $$A$$, let the functor $$\phi_X\colon \int X\to [A^{op},\mathsf{Set}]$$ be the composition of the forgetful functor $$\prod_X\colon \int X\to A$$ and the Yoneda embedding $$Y_A\colon A\to [A^{op},\mathsf{Set}]$$. Define a cocone $$\lambda^X\colon \phi_X\Rightarrow X$$ given by $$(\lambda^X)_{(a,s)} = y^{-1}_{a,X}(s)$$ where $$y_{a,X}\colon\mathsf{Hom}_{[A^{op},\mathsf{Set}]}(\mathsf{Hom}_A(-,a),A)\to A_a$$ is the natural bijection from the Yonede lemma. Then $$\lambda^X$$ is a colimit cocone.

Cisinski, Proposition 1.1.10 Let $$A$$ be a small category, $$C$$ a cocomplete locally small category and $$u\colon A\to C$$ a functor. For each presheaf $$X$$ over $$A$$, define a functor $$u_X\colon \int X\to C$$ given by $$u_X(a,s) = u(a)$$ and $$u_X(f) = u(f)$$. For each presheaf $$X$$, choose a colimit $$L_X$$ and a colimit cocone $$\mu^X\colon u_X\Rightarrow L_X$$. Define a functor $$u_!\colon [A^{op},\mathsf{Set}]\to C$$ by making it send a presheaf $$X$$ to $$L_X$$ and a morphism $$f\colon X\Rightarrow Y$$ of presheaves to the unique morphism $$u_!(f)\colon L_X\to L_Y$$ such that $$u_!(f)\circ\mu^X_{(a,s)} = \mu^Y_{(a,u_a(s))}$$ for any $$(a,s) \in \int X$$. Then $$u_!$$ has a right adjoint (for detalis about this proposition, see this question).

Finally, the remark in question.

Cisinksi, Remark 1.1.11 The functor $$u_!$$ will be called the extension of $$u$$ by colimits. In fact, any cocontinuous functor $$F\colon [A^{op},\mathsf{Set}]\to C$$ is isomorphic to the functor of the form $$u_!$$. More precisely, if we put $$u(a) = F(\mathsf{Hom}_A(-,a))$$ and $$u(f) = F(\mathsf{Hom}_A(-,f))$$, there is a unique natural isomorphism $$u_!(X) \cong F(X)$$ which is the identity whenever the presheaf $$X$$ is representable.

I've constructed a natural isomorphism $$\eta\colon u_!\Rightarrow F$$ by setting $$\eta_X\colon u_!(X)\to F(X)$$ be the unique morphism for which we have $$\eta_X\circ \mu^X_{(a,s)} = F(\lambda^X_{(a,s)})$$ for any $$(a,s) \in \int X$$ ($$\mu^X$$ and $$\lambda^X$$ mean what they meant above).

My question is:

• Is my natural isomorphism gives identity when $$X$$ is representable? If so, why?

• If not, what is the correct natural isomorphism?

• At any case, how to prove uniqueness of said natural isomorphism which gives identity whenever whenever it's value is a representable presheaf?

Your first question doesn't really make sense, or rather it's not reasonable to expect it to be the identity : that would mean you have $$u_!(X) = F(X)$$ with a hard equal, given the definition of $$u_!$$, it's not reasonable.

However, you can indeed note that $$u_!(X)$$ has two isomorphisms with $$F(X)$$ when $$X$$ is representable : the one you constructed here, and the one given in your other question : you can ask whether they're the same (i.e. if you consider the second one to be an identification, then the second one is the identity with respect to that identification - I think that's what Cisinski means)

Indeed, note that for a represented presheaf $$X= \hom(-,b)$$, we have $$\hom( u_!X,Y) \cong \hom(X, u^*Y) \cong u^*Y(b) = \hom(u(b), Y) = \hom(F(\hom(-,b)), Y)$$

The string of isomorphisms up to $$\hom(u(b), Y)$$ yields the iso $$u_!X \cong u(b)$$ that you had defined in your previous question (by definition); and then you can note that since $$(b,id_b)$$ is terminal in $$\int X$$, $$u(b) \to u_!X$$ ($$\mu^X_{(b,id_b)}$$, the canonical inclusion) is an isomorphism, and it suffices to check that this is indeed the same as the one given by our string of isomorphisms.

But to check this, one only needs to remember where the adjunction $$u_! \dashv u^*$$ came from in the first place : precisely from the same type of canonical inclusion. Let's take $$Y= u_!X$$ and follow $$id_{u_!X}$$ : it goes to $$a\mapsto (s\mapsto (u(a)= u_X(a,s) \overset{\mu^X_{(a,s)}}\to u_!X\to u_!X))$$ so to $$a\mapsto (s\mapsto (u(a)= u_X(a,s) \overset{\mu^X_{(a,s)}}\to u_!X))$$;

then you evaluate that in $$b,id_b$$ so you get precisely $$u(b) \overset{\mu^X_{(b,id_b)}}\to u_!X$$.

So the two isomorphisms are the same in the case of a representable presheaf, which is the best we can get if we want $$\eta$$ to be "the identity on representable presheaves".

For your last question, this simply follows from any presheaf being a canonical colimit of representable presheaves : if you have two natural morphisms $$h,k : F\to G$$ between a colimit preserving functor $$F$$ and $$G$$ any functor on $$\widehat{A}$$ which agree on representables, then $$h=k$$.

Indeed, let $$X$$ be any presheaf, we want to check that $$h_X = k_X$$; for that it suffices to show that $$h_X \circ F(\lambda^X_{(a,s)}) = k_X \circ F(\lambda^X_{(a,s)})$$ for any $$(a,s) \in \int X$$ (by the definition of colimit, and because $$F$$ respects them)

But this follows because the LHS is just $$G(\lambda^X_{(a,s)})\circ h_{\hom(-,a)}$$ by naturality, and the RHS is $$G(\lambda^X_{(a,s)})\circ k_{\hom(-,a)}$$ for the same reason, but $$h_{\hom(-,a)} = k_{\hom(-,a)}$$ by assumption, so RHS = LHS, and so $$h=k$$.

Since $$u_!$$ preserves colimits, there is at most one natural iso that agrees with $$\eta$$ on representables : this is the uniqueness statement you wanted

• Dear Max, it seems you are quite proficient at category theory. If you can, please, have a look at this question of mine: math.stackexchange.com/q/3463494/229776 Dec 5, 2019 at 12:22
• We can ask for hard equality $u_!(X)=F(X)$. The point is that $u_!$ is constructed by choosing, for each $X$ a suitable colimit. We may use the axiom of choice for this, but, since $F(X)$ is such a choice for each $X$, we may as well take it. The constructive version of the proof thus gives a genuine equality. Nov 9, 2020 at 9:50