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One of the consequences of the Yoneda embedding is that, given a category $C$ and two objects $A, B$ in $C$, we can obtain an isomorphism between $A$ and $B$ by finding a natural isomorphism between the presheaves $\mathrm{Hom}_C(-, A)$ and $\mathrm{Hom}_C(-, B)$. But suppose we define a smaller category $C^-$ that throws away every object in $C$ besides $A$ and $B$, and throws away every morphism that involves anything besides those two objects. The Yoneda embedding again says that $A$ is isomorphic to $B$ in $C^-$ (which implies isomorphism in $C$) if and only if $\mathrm{Hom}_{C^-}(-, A)$ is naturally isomorphic to $\mathrm{Hom}_{C^-}(-, B)$. And a natural isomorphism between the latter two presheaves seems like it should be strictly easier to obtain, since we have fewer components of the natural isomorphism that we have to define.

In other words, it seems like to show $A$ and $B$ are isomorphic in $C$, we only need to look at “probes” into them from those two objects alone. So what is the advantage of working with all the extra “probes” in the larger category with all the other objects?

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    $\begingroup$ On a typographical note, you probably should make the category $C$ look different from its objects $A$ and $B$. People either use \mathcal, \mathscr, \mathbf, \mathsf or \mathbb for categories (roughly ordered in popularity), or they denote the objects of $C$ by lowercase letters. Worst case, they use letters $X, Y, \dotsc$ that are further away from $C$. $\endgroup$ Commented Jan 3, 2023 at 5:51

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This is against the philosophy of the Yoneda Lemma.

When we want to prove $X \cong Y$ via the Yoneda Lemma, the whole idea is to look at the whole category and compare how $X$ and $Y$ relate to its objects. Theoretically you could, of course, just consider the full subcategory on $\{X,Y\}$, but this restriction brings extra complexity which would make the proof more complicated. This is because we would focus more on specific objects and their specific properties instead of the whole category.

Also, often the proof works by using universal properties of $X$ and $Y$. Typically, they refer to the whole category. So it is not necessary to restrict it.

Let me show this via an example. If $G$ is a group with two normal subgroups $N \subseteq M$, then $(G/N)/(M/N) \cong G/M$. Here is the Yoneda proof, using the universal properties of quotient groups: If $H$ is a group, we have natural bijections

$$\begin{align*} \hom((G/N)/(M/N),H) & \cong \{\varphi \in \hom(G/N,H) : \varphi|_{M/N} = 1 \} \\ & \cong \{\psi \in \hom(G,H) : \psi|_N = 1, \, \psi|_M=1\} \\ & = \{\psi \in \hom(G,H) : \psi|_M=1\} \\ & \cong \hom(G/M,H). \quad \checkmark\end{align*}$$

As you can see, it is absolutely not necessary to restrict $H$ somehow. And if we did, we would probably start wondering how to use the specific properties of $H$. For example, if we only had $H \in \{G/M, (G/N)/(M/N)\}$, we would start wondering how to construct $\hom((G/N)/(M/N),(G/N)/(M/N)) \cong \hom(G/M,(G/N)/(M/N))$, which is not clear at all. We would probably need to find a map $G/M \to (G/N)/(M/N)$, thus proving the whole thing directly and rendering the whole Yoneda approach useless.

There are many more specific applications of the Yoneda Lemma (some of them compiled in my category theory textbook, by the way). You always find the same pattern: the specific choice of the test object ("probe") does not matter.

Related: Generally speaking categories of "nice" objects (for instance, here the subcategory containing only the two objects we are interested in) tend to be badly behaved.

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    $\begingroup$ To summarize: the Yoneda "philosophy" suggests constructing isomorphisms by constructing natural bijections from ready-made natural bijections induced by universal properties. Constructing natural bijections out of components would be unnecessarily complex. This is because the Yoneda lemma asserts that any morphism $A\to B$ is the image of the identity morphism $\mathrm{id}_A\colon A\to A$ under the $A$-indexed component of a unique natural transformation $\mathrm{Hom}_C(-, A)\Rightarrow\mathrm{Hom}_C(-, B)$. So the $A$-indexed component would already contain the data of the isomorphism. $\endgroup$ Commented Jan 3, 2023 at 16:18
  • $\begingroup$ @MartinBrandenburg: Nice answer (and thanks for the reference to your book :-)) (+1) $\endgroup$ Commented Jan 3, 2023 at 20:51

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