"Philosophical" meaning of the Yoneda Lemma - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:52:07Z http://mathoverflow.net/feeds/question/3184 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma "Philosophical" meaning of the Yoneda Lemma Sam Derbyshire 2009-10-29T01:33:30Z 2013-02-25T13:07:02Z <p>The Yoneda Lemma is a simple result of category theory, and its proof is very straightforward.</p> <p>Yet I feel like I do not truly understand what it is about; I have seen a few comments here mentioning how it has deeper implications into how to think about representable functors.</p> <p>What are some examples of this? How should one think of the Yoneda Lemma?</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/3185#3185 Answer by Urs Schreiber for "Philosophical" meaning of the Yoneda Lemma Urs Schreiber 2009-10-29T01:45:38Z 2013-02-12T10:43:15Z <p>One way to look at it is this:</p> <p>for $C$ a category, one wants to look at presheaves on $C$ as being "generalized objects modeled on $C$" in the sense that these are objects that have a sensible rule for how to map objects of $C$ into them. You can "probe" them by test objects in $C$.</p> <p>For that interpretation to be consistent, it must be true that some $X$ in $C$ regarded as just an object of $C$ or regarded as a generalized object is the same thing. Otherwise it is inconsistent to say that presheaves on $C$ are generalized objects on $C$.</p> <p>The Yoneda lemma ensures precisely that this is the case. </p> <p>I wrote up a more detailed expository version of this story at <a href="http://ncatlab.org/nlab/show/motivation+for+sheaves%2C+cohomology+and+higher+stacks" rel="nofollow">motvation for sheaves, cohomology and higher stacks</a>.</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/3199#3199 Answer by Peter Arndt for "Philosophical" meaning of the Yoneda Lemma Peter Arndt 2009-10-29T03:06:43Z 2009-10-29T03:06:43Z <p>A good and frequent use of the Yoneda lemma is internalization: If e.g. I have monoid valued representable contravariant functor Hom(-,A):C-->Set, then the representing object A must be a monoid object in C. This is because the structure morphism Hom(-,A)xHom(-,A)=Hom(-,AxA)-->Hom(-,A) is a natural transformation and thus, by Yoneda comes from a morphism AxA-->A inside C, same for the other structure morphisms and the commuting diagrams.</p> <p>The same goes through for other algebraic (or limit) structures and also for covariant Hom-functors which, if they are algebra-valued are represented by an coalgebra-object. An excellent example for the latter is the fact that affine algebraic groups are represented by Hopf algebras.</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/3206#3206 Answer by Mike Shulman for "Philosophical" meaning of the Yoneda Lemma Mike Shulman 2009-10-29T03:46:25Z 2009-10-29T03:46:25Z <p>Another way to think about the Yoneda lemma is in terms of <em>universal things</em>. Consider, for instance, the existence of classifying spaces for bundles. The statement is that for any suitable group G, there is a space BG such that for any nice enough space X, homotopy classes of maps X → BG are in natural bijection with isomorphism classes of G-structured bundles over X. In categorical terms, that means there is a natural isomorphism between the functors</p> <p>X ↦ {G-structured bundles over X}</p> <p>and</p> <p>X ↦ [X,BG]</p> <p>The Yoneda lemma implies that this natural isomorphism is uniquely determined by a specific G-structured bundle over BG. That is, the existence of a "classifying space" BG with the above property implies the existence of a <em>universal bundle</em> EG → BG such that every bundle over any space X is the pullback of the universal one along a map X → BG, unique up to homotopy.</p> <p>The search for representing objects, and hence for universal data, lies at the heart of a lot of modern algebraic topology, algebraic geometry, and even category theory.</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/3208#3208 Answer by Tom Leinster for "Philosophical" meaning of the Yoneda Lemma Tom Leinster 2009-10-29T04:03:02Z 2013-02-25T13:07:02Z <p>Lazily, I'll just point to some notes on this question: <a href="http://www.maths.ed.ac.uk/~tl/categories/yoneda.ps" rel="nofollow">What's the Yoneda Lemma all about?</a></p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/3223#3223 Answer by Theo Johnson-Freyd for "Philosophical" meaning of the Yoneda Lemma Theo Johnson-Freyd 2009-10-29T06:57:02Z 2009-10-29T06:57:02Z <p>In his Algebraic Geometry class a few years back, Ravi Vakil explained Yoneda's lemma like this: You work at a particle accelerator. You want to understand some particle. All you can do are throw other particles at it and see what happens. If you understand how your mystery particle responds to all possible test particles at all possible test energies, then you know everything there is to know about your mystery particle.</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/3348#3348 Answer by Dan Piponi for "Philosophical" meaning of the Yoneda Lemma Dan Piponi 2009-10-29T22:04:34Z 2009-10-29T22:04:34Z <p>If you don't mind thinking of category theory in terms of functional programming there is an interpretation <a href="http://blog.sigfpe.com/2006/11/yoneda-lemma.html" rel="nofollow">here</a>. Fix a type A and a functor F. If you have a machine that can give you back an object of type FB every time you give it a function of type A->B, can you reverse engineer fully what the machine is doing? Essentially the machine must contain an element of FA and you can recover that FA from how it responds to your functions. This is very similar to Theo's physical perspective.</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/3657#3657 Answer by Aleks Kissinger for "Philosophical" meaning of the Yoneda Lemma Aleks Kissinger 2009-11-01T11:27:09Z 2009-11-01T11:27:09Z <p>Barr and Wells (Toposes, Triples, and Theories, 84) talks about arrows as a general kind of elements. In Set, arrows from {*}&rarr;A are the usual elements of A, and arrows from bigger sets X&rarr;A are the X-elements of A, or elements of A parameterised in X. Of course the latter makes sense in any category, so we can use this language the state the Yoneda lemma as:</p> <p>The Hom(-,A)-elements of F are just the usual elements of FA.</p> <p>I find this to be, at least, a useful mnemonic, but also justifies the intuition that an object "is" its collection of probes.</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/15143#15143 Answer by Sridhar Ramesh for "Philosophical" meaning of the Yoneda Lemma Sridhar Ramesh 2010-02-12T21:58:05Z 2010-02-13T07:45:52Z <p>If you have basic experience with abstract algebra, the ideas in the Yoneda lemma should be quite familiar and even intuitive; the apparent difficulty is only in recognizing them in this new presentation.</p> <p>You can think of "category" as meaning the same thing as "algebraic theory in a multisorted language with only unary functions" (the objects of the category being the sorts of the language, the morphisms being the definable functions, and the equalities between (composites of) morphisms being the laws of the theory). From this perspective, a functor from C to Set is simply a model of the theory corresponding to C, and natural transformations of such functors are homomorphisms of models. The Yoneda lemma then is about free models: specifically, it says that for every sort s, the "term model" of terms with a single variable, of sort s (equivalently, definable functions with domain s) is the free model on a single generator of sort s. [It may be unfamiliar when expressed as "Nat(Hom(s, -), M) ~= M(s), naturally in M", but that is indeed all this categorical expression is saying]</p> <p>The so-called co-Yoneda lemma mentioned in the other comments also has a nice interpretation from this perspective, amounting to the demonstration that every model can be specified by generators and relations.</p> <p>(I wouldn't say this is The One Right Way to think about the Yoneda lemma, because it's useful to view it from many different perspectives, but this is certainly One Right Way to think about the Yoneda lemma.)</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/121397#121397 Answer by Daniel Moskovich for "Philosophical" meaning of the Yoneda Lemma Daniel Moskovich 2013-02-10T15:11:13Z 2013-02-12T08:37:31Z <p>You might also want to think about the Yoneda Lemma as a statement about functors.</p> <p>A locally small category $\mathcal{C}$ is embedded by the $\mathrm{hom}$ functor in the category $\mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathrm{set})$. This is called the <em>Yoneda embedding</em>. Thus, the $\mathrm{hom}$ functor is fully faithful (this itself is a corollary of the Yoneda lemma), but is <b>not</b> an equivalence of categories, because it isn't essentially surjective. In other words, not every functor $F$ from $\mathcal{C}^{\mathrm{op}}$ to $\mathrm{set}$ is representable- for example, the empty functor which maps each object in $\mathcal{C}$ to the empty set, is never representable. The problem is that the Yoneda embedding does not commute with colimits. But the Yoneda lemma tells you that every functor $F\in \mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathrm{set})$ becomes representable when extended appropriately. In other words, every functor $F$ from $\mathcal{C}^{\mathrm{op}}$ to $\mathrm{set}$ <b>extends</b> to a functor from $\left(\mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathrm{set})\right)^{\mathrm{op}}$ to $\mathrm{set}$ (this is a special case of the <em>Yoneda extension</em>) which does commute with colimits, and is representable. </p> <p>So one philosophical interpretation'' of the Yoneda lemma is the following:</p> <blockquote> Every functor $F$ from $\mathcal{C}^{\mathrm{op}}$ to $\mathrm{set}$ can be extended to a representable functor from $\left(\mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathrm{set})\right)^{\mathrm{op}}$ to $\mathrm{set}$. </blockquote> <p>One reference for this point of view is <a href="http://people.fas.harvard.edu/~amathew/doldkan.pdf" rel="nofollow">these notes by Akhil Mathew</a>.</p> http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/121410#121410 Answer by Kai for "Philosophical" meaning of the Yoneda Lemma Kai 2013-02-10T17:57:41Z 2013-02-12T09:12:41Z <p>Here is an example on representable functors. Yoneda's lemma gives down-to-earth, morpshim oriented interpretation of representable functors, and vice versa.</p> <p>I will explain this with an example.</p> <blockquote> <p>In a category $\mathscr{C}$, the product of $A$ and $B$ is the pair of object $A\times B$ in $\mathscr{C}$ and a fixed natural isomorphism $$\sigma \colon \mathrm{Hom}(-,A\times B)\to \mathrm{Hom}(-,A)\times \mathrm{Hom}(-,B).$$</p> </blockquote> <p>This definition of products only uses terminology of functors. By applying Yoneda's lemma, we arrive at a morphism oriented definiton of products. Yoneda's lemma says that there is a bijection $$\Psi \colon \mathrm{Hom}\left( \mathrm{Hom}(-,A\times B),\mathrm{Hom}(-,A)\times \mathrm{Hom}(-,B)\right) \to \mathrm{Hom}(A\times B,A)\times \mathrm{Hom}(A\times B,B).$$ In particular, we apply this to $\sigma$ and denote $$\Psi(\sigma)=\sigma(A\times B)(\mathrm{id}_{A\times B})=(\pi^{A}\colon A\times B\to A,\pi^{B}\colon A\times B\to B).$$ Next, by applying the inverse of $\Psi$, we compute $$\sigma(X)=\Psi^{-1}\left( \Psi(\sigma)\right)(X):\mathrm{Hom}(X,A\times B)\to \mathrm{Hom}(X,A)\times \mathrm{Hom}(X,B)$$ $$f\colon X\to A\times B\mapsto (\pi^{A}\circ f,\pi^{B}\circ f).$$ Since $\sigma$ is a natural isomorphism, $\sigma(X)$ is a bijection. This bijectivity is the usual definition of product based on morphisms (universality):</p> <blockquote> <p>For any pair of morphisms $f^{A}\colon X\to A$ and $f^{B}\colon X\to B$, there exists a unique morphism $f\colon X\to A\times B$ with $\pi^{A}\circ f=f^{A}$ and $\pi^{B}\circ f=f^{B}$.</p> </blockquote> <p>I think the Philosophy behind Yoneda's lemma is that, it connects the world of functors (and natural transformations) $\mathfrak{Set}^{\mathscr{C}^{\mathrm{op}}}$ and the world of morphisms $\mathscr{C}$.</p>