Why presheaves are generalized objects?

Question

While self studying category theory (Yoneda lemma), I came across the statement that for any category $\mathsf{C}$ the functor category $\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set})$ represents generalized objects of $\mathsf{C}.$

Here generalized means bunch of objects of $\mathsf{C}$ glued together.

Because of the Yoneda embedding $$Y:\mathsf{C}\to\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set}),$$ I can imagine that $\mathsf{C}$ lives inside of $\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set})$ as $Y(\mathsf{C}),$ however I can not see why other objects in this category acts like generalized objects of $\mathsf{C}.$

Can anybody explain me why this philosophy works, possibly with some example.

Answer 1

There are several ways of seeing this. The Yoneda embedding tells you to treat each object of $C$ as the constant presheaf. Much like you can think of a real number as a constant sequence of real numbers. Now, if you allow more variation in the sequence of numbers, but still insist on using real numbers, then you can think of an arbitrary sequence as a generalized real number. But, you can get really crazy wild sequences like that and it is questionable whether they should be considered as generalised real numbers. So, change to a more familiar scenario: sequences of rational numbers. Here we can use the Cauchy condition to tame our sequences and stay close to the original rationals. So, we can think of Cauchy sequences of rational numbers as generalized rational numbers. Taking a quotient of those and we end up with the reals. So, we can think of the reals as being generalized rationals. More precisely, the reals are obtained as a completion in this way: we have our original rationals viewed as constant sequences, we've added more general sequences (with some equivalence relation, but don't mind that) and what we got in the end of not much larger in the sense that every bounded above set of rationals now has a supremum and vice versa each new element is the supremum of a bunch of rationals.

Now, the presheaf category has a similar property: Every presheaf is a colimit of representables, namely the Yoneda embedding, much like viewing a number as a constant sequence, allows us to reach each presheaf as a colimit of things in its image. This analogy goes deeper than that when you consider the enriched Yoneda in the context of generalized metric spaces (Lawvere spaces).

Answer 2

Here's a proof of the property that Ittay Weiss alluded to and that was mentioned in the comments :

Let $\newcommand{\C}{\mathsf C} \newcommand{\set}{\mathsf{Set}} \newcommand{\y}{\mathsf Y}\newcommand{\fun}{\mathsf{Fun}} F:\C^{op}\to \set $ be a functor and let $\int_{\C} F$ be the following category : its objects are couples $(x,s)$ where $x$ is an object of $\C$ and $s\in F(x)$, and a morphism $(x,s)\to (y,t)$ is a morphism $f:x\to y \in \C$ such that $F(f)(t) = s$ (it makes sense as $F$ is contravariant on $\C$). Composition and identities are defined the obvious way.

Then you have a projection $\int_\C F\to \C$ defined as $(x,s)\to (y,t) \mapsto x\to y$. This is clearly a functor. The claim is that $\int_C F\to\C \to \fun(\C^{op},\set)$ has $F$ as a colimit.

To understand why this construction makes sense first, it'd be good for you to see how it relates to the comma category $\fun(\C^{op},\set)/F$ (hint : it should be the full subcategory of $\fun(\C^{op},\set)/F$ on representable presheaves: we're taking all morphisms $\y (c) \to F$ and their colimit should be $F$, that makes intuitive sense)

Now for the proof, I could write it out, but it's full of details and it's mainly the Yoneda lemma at all stages. I do recommend you try it out for yourself now that you have the specific info.

If you don't manage to do it, you can look up my answer there. The notations of the questions and the formulation aren't exactly the same, but it's the same theorem that's being proved and in my answer there I used notations closer to the ones I introduced here (if you have trouble translating the question there and relating it to my claim here, you are of course welcome to ask for more)

Answer 3

The previous answers are very good, but I also like to always keep in mind a simple example when working with presheaves, to get a feel for what all this means.

Luckily, we have a very simple and intuitive category of presheaves to wrok with. Consider the category $\mathbb{G}$, whose objects are $[0]$ and $[1]$, and whose non identity morphisms are $\sigma,\tau : [0] \to [1]$. A preseaf $X$ over $\mathbb{G}$ is giben by two sets $X_{[0]}, X_{[1]}$, together with two applications $s,t : X_{[1]}\to X_{[0]}$. You might recognise from this the definition of a graph (or maybe you call it a multigraph, since you can always have multiple egdes between two vertices, but I will call these graphs in the following). Explicitly, $X_{[0]}$ is the set of vertices, $X_{[1]}$ the set of edges, $s$ associates to each edge its source, and $t$ associates to each edge its target.

You can work out that the representable $Y([0])$ is actually the graph consisting of a single point, and $Y([1])$ is the graph consisting of a single arrow between two different points.

Rephrasing your statement for this special cases reads "graphs are a generalisation of points and arrows". I find this quite enlightening, to understand what "glued together" means and how the original statement should be understood

Answer 4

This may not be technical as other answers. But I haven't seen this very intuitive explanation written anywhere, and therefore wanted to share with others. Suppose we have a category $\mathsf{C}$ that we want to somehow (sorry, this is too vague) generalize without adding more morphisms, but adding more objects. Lets say this new (unknown generalized) category is $\tilde{\mathsf{C}},$ which comes with a fully faithful functor $\iota: \mathsf{C}\hookrightarrow\tilde{\mathsf{C}}$. For any generalized object $A\in\tilde{\mathsf{C}},$ we have a functor $$Y_A:\mathsf{C}^{op}\to\mathsf{Set}$$ given by $Y_A(X)=\mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X, A).$ This functor is contravariant (or covariant according to my notation) because any morphism $X\xrightarrow{f} X'$ in $\mathsf{C}$ induces a function $Y_A(X')\xrightarrow{(\_)\circ\iota f}Y_A(X)$ by precomposition. Now we can implement this construction to another functor $$Y: \tilde{\mathsf{C}}\to\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set})$$ such that $Y(A)=Y_A.$ For any morpsism $A\xrightarrow{g}A'$ in $\tilde{\mathsf{C}}$ we have a natural transformation $Y(A)\Rightarrow Y(A')$ whose components are given in the commutative square $\require{AMScd}$ \begin{CD} \mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X', A) @>(\_)\circ f>> \mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X, A)\\ @V g\circ(\_) V V @VV g\circ(\_) V\\ \mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X', A') @>>(\_)\circ f> \mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X, A') \end{CD}

By the Yoneda lemma $Y\vert_{\mathsf{C}}$ is fully faithful, therefore contains all the information of $\mathsf{C}$ (and possibly more interesting objects in the restricted target). The most important part of this seemingly trivial construction is to realize that $Y\vert_{\mathsf{C}}$ has no reference to its generalization $\tilde{\mathsf{C}}.$ So, we declare that unknown category $\tilde{\mathsf{C}}$ to be the category of preshaves $\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set}).$ In other words, $Y$ is an isomorphism of categories and $Y\vert_{\mathsf{C}}$ is the Yoneda embedding.

Next, the co-Yoneda lemma says that we haven't added any wild objects while generalizing $\mathsf{C}$ in this manner. In technical terms it say every presheaf is a colimit of objects in the image of $Y\vert_{\mathsf{C}}$ (this image is called the representable presheaves). Also, the category of preshaves is cocomplete. So, we think of it as the free cocompletion of $\mathsf{C}.$