Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Presheaves are not the free completion of the underlying cat

Peva Blanchard (Jun 24 2024 at 12:05):

Given a category $C$, it is known that the category of presheaves $\text{Psh}(C) = [C^{op}, \text{Set}]$

• is the free co-completion of $C$
• and is also complete

But, is there a specific example that shows that $\text{Psh}(C)$ is not the free completion of $C$?

John Baez (Jun 24 2024 at 12:19):

$C = 1$ should do it. If you believe the free cocompletion of the terminal category is $\mathsf{Set}$, you should also believe the free completion of the terminal category is $\mathsf{Set}^{\rm{op}}$, so all you need to show is that $\mathsf{Set}^{\rm{op}}$ is not equivalent to $\mathsf{Set}$. For this, you can note that binary products distribute over binary coproducts in $\mathsf{Set}$, but not the other way around.

Peva Blanchard (Jun 24 2024 at 12:21):

Thank you!

Tom Hirschowitz (Jun 25 2024 at 07:31):

Neat argument!
Maybe a useful perspective on the asymmetry of the situation is that the Yoneda embedding $C \to 𝐏𝐬𝐡(C)$ is continuous but not cocontinuous. So the presheaf construction freely adds colimits, but adds limits non-freely, i.e., preserving existing ones.

John Baez (Jun 25 2024 at 09:36):

Yes, that's nice. And reverting to the example $C = 1$, we can clearly see the Yoneda embedding $1 \to \mathsf{Set}$ is preserving products but not coproducts, since in $\mathsf{Set}$ we have $1 \times 1 \cong 1$ but sadly not $1 + 1 \cong 1$.

Valeria de Paiva (Jun 25 2024 at 09:43):

since in Set we have 1×1≅1 but sadly not 1+1≅1.

well, I do not think this is "sadly", as the fact that true and false are different is logically very important!

John Baez (Jun 25 2024 at 09:44):

I was joking, but: life would be so much simpler the Yoneda embedding preserved limits and colimits, so that 1+1≅1, and true = false.

John Baez (Jun 25 2024 at 09:45):

By the way that's an interesting form of proof by contradiction: actually concluding that true = false.

Peva Blanchard (Jun 25 2024 at 10:47):

I often think of "non-free object" as something like "free object + non-trivial equations", like the presentation of a group.
If I understand correctly, is it correct to say that the equation $1 \times 1 \cong 1$ in $\text{Set}$ is one such equation?

Rémy Tuyéras (Jun 25 2024 at 11:38):

The problem with "finding" a free completion in $[C^{op},\mathbf{Set}]$ is the morphisms. For example, we could ask whether the restriction of $[C^{op},\mathbf{Set}]$ to the generators (the Yoneda embeddings) and all their limits is a free completion? This will generally not happen because we will struggle to describe the morphisms of the form

$\mathsf{lim}_iY_{a_i} \to \mathsf{lim}_jY_{b_j}$

freely. On the other hand, using Yoneda Lemma, the morphisms of the form

$\mathsf{colim}_iY_{a_i} \to \mathsf{colim}_jY_{b_j}$

can be described by the set $\mathsf{lim}_i\mathsf{colim}_jC(a_i,b_j)$.

John Baez (Jun 25 2024 at 11:49):

Peva Blanchard said:

I often think of "non-free object" as something like "free object + non-trivial equations", like the presentation of a group.
If I understand correctly, is it correct to say that the equation $1 \times 1 \cong 1$ in $\text{Set}$ is one such equation?

This is a somewhat confusing question because there's another way something can fail to be "free on X": it can have a bunch of extra stuff. And depending on what you're doing, that may be happening here.

We don't get $\mathsf{Set}$ simply by first taking the free completion of the terminal category and then imposing extra equations (isomorphisms): from this viewpoint it also has a lot of extra objects that didn't arise as limits from the first object you put in, the object coming from the terminal category, which I assume you are calling $1$.

But anyway, if we were trying to form the free category with limits on one object $x$, we would not make $x$ be terminal, so we would not have $x \times x \cong x$. So maybe the answer to your question is "yes".

Rémy Tuyéras (Jun 25 2024 at 13:02):

@Peva Blanchard To align my comment with @John Baez 's comment and your question, the equation $x \cong x \times x$ (or $1 \times 1 \cong 1$) forces the two projections $x \times x \to x$ to be equal (to a certain isomorphism). This constitutes a non-trivial equation in the homset $\mathsf{hom}(x \times x, x)$