Abstract Nonsense

Crushing one theorem at a time

Yoneda’s Lemma (Pt. I)


Point of Post: In this post we discuss the Yoneda lemma, the Yoneda embedding, and associated corollaries.

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Motivation

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Yoneda’s lemma is one of those theorems that pop up periodically in mathematics that are best described as “easy to understand, hard to get.” By this, I mean that it’s really not too hard (assuming your comfortable with categories) to understand the statement of Yoneda’s lemma, and really not any more difficult to understand the proof. That said, it’s hard to ‘get’ what Yoneda’s lemma says, and why it is so very important. So, what is this lemma? The basic premise of Yoneda’s lemma starts with a simple sort of observation about categories. Namely, there are technically two ingredients to defining a category: the objects and the arrows. That said, as one progresses on their categorical journey through mathematical life one begins to realize a startling fact–the objects aren’t important. Well, to be more precise, the objects really serve mainly as a convenient way to index arrows, which tells us when we can compose two arrows.  Intuitively then we should be able to consider the arrows h_A in category ending at some object A as being some sort of ‘idealized representation’ of A. Of course, for this intuition to have any sort of rigorous grounding we’d like this representation to be “faithful” in the sense that we’d like to be able to distinguish between different objects by their corresponding set of arrows. Of course, doing category theory where we really only care about things up to isomorphism what we really mean is that we’d like to be able to tell non-isomorphic things apart by looking at their corresponding set of arrows. In a very rough sense Yoneda’s lemma tells us that this, in fact, true. Namely, that if h_A\cong h_B (whatever that’s supposed to mean) that A\cong B.

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That was all easy, right? The statement of Yoneda’s lemma is really not that astounding sounding at first, but this precisely the trap that we alluded to. So, what does Yoneda’s lemma really help us with? Stealing a turn of phrase from Tom Leinster one can conceptulaize \text{Hom}(A,B) as “B‘s view from A“, and then Yoneda’s lemma can be thought of as saying “If X and Y look the same from every vantage point, they must be isomorphic.” So what though? That’s cool and all, but is it really that profound. Yes, yes it is. In fact, at the risk of sounding dramatic, Yoneda’s lemma provides a uniform methodological approach to studying any class of mathematical objects. Indeed, Yoneda’s lemma tells us that if we want to know the true character of an object X we should study \text{Hom}(A,X) for all the objects A we can muster. Intuitively, each object A “probes” X via \text{Hom}(A,X) adding a small piece of the puzzle to the overall tapestry that is h_X, which tells us “something” about X. For example, the fact that \text{Hom}_\mathbb{Z}(\mathbb{R},\mathbb{Z})=\{0\} really gives a pretty important group theoretic (partial) picture of \mathbb{R}.

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Of course, it should be conceptually clear that if we can completely characterize (up to isomorphism) objects by looking at the morphisms into them then we should also be able to characterize the objects by the set h^A of morphisms out of them.

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So, let’s see if we can use the above to see why it makes sense that certain subjects are so powerful. To this end, let’s think about something for a second. Suppose that an object X sits isomorphically inside some object Y (think of the more common, concrete categories, if it helps at this point). Then, studying \text{Hom}(A,X) and \text{Hom}(A,Y)  is somewhat redundant since \text{Hom}(A,X)\hookrightarrow\text{Hom}(A,Y) so that intuitively \text{Hom}(A,Y) contains the information of \text{Hom}(A,X)–using our metaphor, the piece of the puzzle corresponding to \text{Hom}(A,Y) sits on top of \text{Hom}(A,X) making \text{Hom}(A,X) useless in the pursuit of “covering” the puzzle. More precisely, this tells us that if we can find a class of objects \{X_\alpha\} for which all objects embed into one of these X_\alpha then to understand an object A it really suffices to look at \text{Hom}(A,X_\alpha) for all \alpha. So, let’s consider some classic examples where a there is a ‘representative’ class of objects, for which all objects embed into. Probably the first that should come to mind is Cayley’s theorem which tells us that every group embeds into S_X for some set X. Thus, to find out the group theoretic properties of some group G the above discussion seems to indicate we can learn a lot by studying \text{Hom}_\mathbf{Grp}(G,S_X) for all sets X–of course, this is definitely true as is extolled in the well learned notion of group actions. But, permutation groups are a little bare, no? It would be preferable if we could somehow find a class of groups in which all groups embed into that have “extra structure” so we could use that extra structure to study the Hom groups. The answer, for finite groups, lies in the fact that S_n\hookrightarrow\text{GL}_n(\mathbb{C}) as permutation matrices so that from the above we could benefit greatly from studying \text{Hom}_{\mathbf{Grp}}(G,\text{GL}_n(\mathbb{C})) for all n–of course, this is wholly true as is seen in the beautiful and useful study of (complex) representation theory of finite groups. As a last example, let’s consider unital rings. We know then that every ring embeds into the endomorphism group of an abelian group. Thus, the above description suggests that to understand a unital ring R we should look at unital ring homomorphisms R\to\text{End}_\mathbb{Z}(A) for some abelian group A–of course, this is the same thing as saying to understand a unital ring R we should look at the modules over R. Once you understand Yoneda’s lemma, and its consequences the reason we study a lot of the things we do, makes a lot more sense.

\text{ } Yoneda’s Lemma

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Ok, so how are we going to make the above formal? The key idea is that instead of actually considering the set of all arrows in or out of some given object we should consider the covariant and contravariant Hom functors associated to that object which, in essence, is the set of all arrows only in “function form”. In particular, if \mathcal{C} is a category we can consider the functor category \mathbf{Set}^{\mathcal{C}^\text{op}} which we shall denote by \mathcal{C}^\wedge. What we then want to consider is an association \mathcal{C}\to\mathcal{C}^\wedge which sends an object x to it’s associated contravariant Hom functor \text{Hom}(\bullet,x). What we’d like to do is make this into a functor F somehow. Namely, we want to figure out a way to associate an arrow x\xrightarrow{f}y to a natural transformation F(f):\text{Hom}(\bullet,x)\implies\text{Hom}(\bullet,y), but how? Namely, given an object z in \mathcal{C} we need to figure out how to create an arrow F(f)(z):\text{Hom}(z,x)\to\text{Hom}(z,y), but it’s obvious how! Namely, we do the old standby by defining F(f)(z) to be the mapping \text{Hom}(z,x)\to\text{Hom}(z,y) given by F(f)(z)(g)=f\circ g. The only thing to check is that this satisfies the naturality condition in z. In other words, if z'\xrightarrow{k^\text{op}}z is some arrow in \mathcal{C}^\text{op} where k is some arrow z\to z' in \mathcal{C} we need to verify that the following diagram commutes

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\begin{matrix}\text{Hom}(z',x) & \overset{\text{Hom}(k^{\text{op}},x)}{\longrightarrow} & \text{Hom}(z,x)\\ ^{F(f)(z')}\big\downarrow & & \big\downarrow ^{F(f(z))}\\ \text{Hom}(z',y) & \underset{\text{Hom}(k^{\text{op}},y)}{\longrightarrow} & \text{Hom}(z,y)\end{matrix}

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To do this, let t\in\text{Hom}(z',x) then we have that

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F(f)(z)(\text{Hom}(k^\text{op},x)(h))=F(f)(z)(t\circ k)=f\circ t\circ k

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and

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\text{Hom}(k^\text{op},y)(F(f)(z')(t))=\text{Hom}(k^\text{op},y)(f\circ t)=f\circ t\circ k.

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Thus, we see that F(f), as defined above, really is a natural transformation \text{Hom}(\bullet,x)\implies\text{Hom}(\bullet,y).  So, we have a mapping \mathcal{C}\to\mathcal{C}^\wedge on both objects and arrows, so let’s verify that it’s actually functorial. But, this really entails checking that F(g\circ f)(z)=F(g)(z)\circ F(f)(z), but this is clear since F(g\circ f)(z)(t)=(g\circ f)\circ t=g\circ(f\circ t)=F(g)(z)(f\circ t)=F(g)(z)(F(f)(z)(t)). Thus, we see that F splits across compositions, and so it remains to check that F respects identities. But, this is clear since F(1_x)(z)(t)=1_x\circ t=t.

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From the above we can really conclude that the association \mathcal{C}\to\mathcal{C}^\wedge defined above is a functor. It’s so important that it’s given a special name, and special symbol. Indeed, instead of F we denote this functor by \mathfrak{Y}^\wedge (if you’re curious, that’s not a weird eta, it’s a capital Fraktur Y) and call it the covariant Yoneda embedding. Wait, what’s an embedding? Intuitively an embedding is a functor which respects the domain category to an extreme degree, in the sense that there is zero “distortion” caused by the functor. Formally, a functor is called an embedding if it’s full, faithful, and injective on objects. The importance of Yoneda’s lemma, as we’re about to discuss, is the theorem (somewhat confusing, considering the name) that the Yonedea embedding really is an embedding.

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References:

[1] Mac, Lane Saunders. Categories for the Working Mathematician. New York: Springer-Verlag, 1994. Print.

[2] Adámek, Jirí, Horst Herrlich, and George E. Strecker. Abstract and Concrete Categories: the Joy of Cats. New York: John Wiley & Sons, 1990. Print.

[3] Berrick, A. J., and M. E. Keating. Categories and Modules with K-theory in View. Cambridge, UK: Cambridge UP, 2000. Print.

[4] Freyd, Peter J. Abelian Categories. New York: Harper & Row, 1964. Print.

[5] Rotman, Joseph J. Introduction to Homological Algebra. Springer-Verlag. Print.

[6] Herrlich, Horst, and George E. Strecker. Category Theory: An Introduction. Lemgo: Heldermann, 2007. Print.

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January 14, 2012 - Posted by | Algebra, Uncategorized | , , , , , , ,

5 Comments »

  1. […] Yoneda Lemma (Pt. II) Point of Post: This is a continuation of this post. […]

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  2. […] the covariant Hom functors associated to and are naturally isomorphic, and then appeal to Yoneda’s lemma to conclude that they are naturally […]

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  3. […] particular, we like to compute the homomorphism group/module if not for practical reasons then for purely philosophical ones. Thus, in this post we shall prove that the Hom functor itself has a certain kind of ‘partial […]

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  4. […] suppose for a second that was such an object. What we have then is a natural isomorphism . But, Yoneda’s lemma tells us that we can associate to this an element defined by . But, of course we see that is […]

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  5. […] so by Yoneda’s lemma . While the basic idea is neat it does not actually prove continuity. I am not saying that one […]

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