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Stream: theory: category theory

Topic: Faithful presheaf over presheaves

Paolo Perrone (Apr 21 2024 at 11:22):

Let $F$ be a presheaf on a category $\mathrm{C}$ (we can assume $$\cat{C}$$ small, if it helps).
Suppose that $F$, as a functor $\mathrm{C}^\mathrm{op}\to\mathrm{Set}$ is faithful.

Consider now the representable presheaf on the category of presheaves $[\mathrm{C}^\mathrm{op},\mathrm{Set}](-,F)$. Is this a faithful functor as well, on $[\mathrm{C}^\mathrm{op},\mathrm{Set}]$?
(Note that, by the Yoneda lemma, its restriction along the Yoneda embedding is $F$, which is faithful.)

Peva Blanchard (Apr 21 2024 at 13:34):

Let $C = 1$ the category with a single object. Then a presheaf on $C$ is just a set, and is necessarily faithful. I choose $F$ to be the singleton set $\{0\}$.

Now, let $G = \{0,1\}$ be the two-element set considered as a (faithful) presheaf on $C$.

• Let $\phi$ be the identity function on $G$ considered as an element of $[C^{op}, Set](G, G)$
• Let $\psi$ be the "swap" function on $G$, also considered as an element of $[C^{op}, Set](G, G)$
  Then $\phi$ and $\psi$ are different.

Pick any $\eta : [C^{op}, Set](G, F)$. There is actually not much choice here: this natural transformation is equivalently described as the unique function $\{0, 1\} \rightarrow \{0\}$. We have in particular:

$$\eta \circ \phi = \eta \circ \psi$$

I think this shows that the representable $[C^{op}, Set](\_, F)$ is not faithful.

Paolo Perrone (Apr 21 2024 at 13:52):

Oh, very nice. Thank you.

Peva Blanchard (Apr 21 2024 at 14:00):

You welcome :) I think though the counter-example is unfair to the problem, because there are no non-trivial morphisms in $C = 1$, so faithfulness comes for free. We should assume $C$ has at least one non identity morphism, so we cannot cheat by choosing $F$ to be the singleton set.

Peva Blanchard (Apr 21 2024 at 21:55):

Mmh, it does not help either.

Take for instance $C$ the category freely generated by the graph with one object $\star$ and one non-trivial morphism $s : \star \rightarrow \star$.

A presheaf $F$ on $C$ is just a set $X$ with a function $f : X \rightarrow X$. The presheaf is faithful iff $f \ne id$. Let's choose $F = \mathbb{N}$ with $f(n) = n + 1$.

Now, let $G$ be another presheaf on $C$, given by a set $Y$ and a function $g : Y \rightarrow Y$.

Assume there exists a natural transformation $\eta : \hat{C}(G, F)$. This means that we have a function $h : Y \rightarrow \mathbb{N}$ s.t. $h(g(y)) = h(y) + 1$ for all $y$. This implies, in particular, that $Y$ is the empty set or infinite.

Stated otherwise, if $Y$ is a non-empty finite set, then $\hat{C}(G, F) = \emptyset$. In that case, any $\psi : \hat{C}(G, G)$ satisfies

$$\forall \eta: \hat{C}(G, F), \eta \circ \psi = \eta \circ id$$

Thus, it suffices to find a $\psi \ne id$ to prove that the representable $\hat{C}(\_, F)$ is not faithful.

Consider the presheaf $G$ with $Y = \{0, 1\}$ and $g = \text{swap}$. Let $\psi = \text{swap}$. This choice defines a natural transformation in $\hat{C}(G, G)$ since $\psi$ commutes with $g$. And $\psi \ne id$.

Therefore $\hat{C}(\_, F)$ is not faithful.

Paolo Perrone (Apr 23 2024 at 18:32):

Thank you.

Paolo Perrone (Apr 23 2024 at 19:03):

Here's a similar question.
An object $S$ of a category $\mathrm{C}$ is called a separator if the functor $\mathrm{C}(S,-):\mathrm{C}\to\mathrm{Set}$ is faithful.
Given such $S$, is the corresponding presheaf $\mathrm{C}(-,S):\mathrm{C}^\mathrm{op}\to\mathrm{Set}$ itself a separator of the category of presheaves?

Peva Blanchard (Apr 25 2024 at 07:33):

This one is harder. I couldn't find the answer, but here is how I approached it.

Let's work in $Prof$.

Let $U_C : C \nrightarrow C$ be the identity profunctor on $C$.
Let $P_S : C \nrightarrow C$ be the profunctor given by

$$P_S(A, B) = Set(C(S, A), C(S, B))$$

We have a 2-cell $\eta_S : U_C \Rightarrow P_S$ (natural transformation given by post-composition $f \mapsto f \circ \_$).
The object $S$ is a separator object when $\eta_S$ is "injective": every component is an injective function. (I'm not sure if this is equivalent to saying that $\eta_S$ is monic).

Let $Y : C \rightarrow \hat{C}$ the Yoneda embedding, seen as a vertical arrow.

Let $\hat{P}_S : \hat{C} \nrightarrow \hat{C}$ be the profunctor given by

$$\begin{align*} \hat{P}_S(F, G) &= Set(\hat{C}(YS, F), \hat{C}(YS, G)) \\ &\cong Set(FS, GS)~\text{(by Yoneda)} \end{align*}$$

We also have a 2-cell $\hat{\eta}_S : U_{\hat{C}} \Rightarrow \hat{P}_S$ given by post-composition.
Using Yoneda, I think $\hat{\eta}_S$ is equivalently described as:

$$\begin{align*} \hat{C}(F, G) &\xrightarrow{\hat{\eta}_{S, FG}} Set(FS, GS) \\ \alpha &\mapsto \alpha_S \end{align*}$$

i.e., it maps any natural transformation to its component at $S$.

$YS$ is a separator in $\hat{C}$ iff $\hat{\eta}_S$ is "injective".
This amounts to say that a natural transformation $\alpha : F \Rightarrow G$ is uniquely determined by its component at $S$.

This seems like it is asking too much, but I don't know exactly where to go from there. I'll continue thinking about it. (It's also possible that I lack experience in dealing with those concepts)

Peva Blanchard (Apr 25 2024 at 11:28):

I think I found a counter-example.

image.png

Let $C$ be the full sub-category of $Set$ spanned by the sets $S = \{0\}$ and $U = \{0,1\}$, depicted on the left in the picture above. If I am not mistaken, $S$ is a separator in $C$.

Let $F, G$ be presheaves on $C$ defined as follows.

$$\begin{align*} FU &= GU = U \\ FS &= GS = S \\ Fv_0 &= Fv_1 = \text{unique map } \{0,1\} \rightarrow \{0\} \\ F! &= \text{constant } 0\\ Gv_0 &= Gv_1 = \text{unique map } \{0,1\} \rightarrow \{0\} \\ G! &= \text{constant } 1\\ \end{align*}$$

There are two distinct natural transformations $F \Rightarrow G$ with the same component at $S$. Their component at $U$ is $\text{swap}$ or the constant $1$ respectively.

This shows that $YS$ is not a separator in $\hat{C}$.

Paolo Perrone (Apr 29 2024 at 10:34):

Thank you!

Reid Barton (May 02 2024 at 17:17):

Kind of the same example, but $[0]$ is a separator in the simplex category $\Delta$, but the corresponding representable $\Delta^0$ is not a separator (homming out of $\Delta^0$ is taking the 0-simplices, and a map of simplicial sets is not determined by its action on 0-simplices).