Yoneda lemma (Rev #3, changes)

Showing changes from revision #2 to #3:
Added | ~~Removed~~ | ~~Chan~~ged

By Lemma 9.5.3 (see product precategory), we have an induced functor $\mathbf{y} : A \to \mathit{Set}^{A^{op}}$ which we call the **yoneda embedding**.

For any precategory $A$, any $a:A$, and any functor $F: \mathit{Set}^{A^{op}}$, we have an isomorphism?

$hom_{\mathit{Set}^{A^{op}}}(\mathbf{y}a,F) \cong F a \qquad \qquad(9.5.5)$

Moreover this is natural in both $a$ and $F$.

**Proof.** Given a natural transformation $\alpha : \mathbf{y}a \to F$, we can consider the component $\alpha_a : \mathbf{y}a(a) \to F a$. Since $\mathbf{y} a(a)\equiv hom_A(a,a)$, we have $1_a:\mathbf{y}a(a)$, so that $\alpha_a(1_a):F a$. This gives a function $\alpha \mapsto \alpha_a(1_a)$ from left to right in (9.5.5).

In the other direction, given $x: F a$, we define $\alpha : \mathbf{y} a \to F$ by

$\alpha_{a'}(f) \eqiv F_{a,a'}(f)(x)$

Naturality is easy to check, so this gives a function from right to left in (9.5.5).

To show that these are inverses, first suppose given $x: F a$. Then with $\alpha$ defined as above, we have $\alpha : \mathbf{y}a \to F$ and define $x$ as above, then for any $f:hom_A(a',a)$ we have

$$\begin{array}{rl}{\mathrm{alpha\alpha}}_{a\prime}(f)& ={\alpha}_{a\prime}(y{a}_{a,a\prime}(f)({1}_{a}))\\ & =({\alpha}_{a\prime}\circ y{a}_{a,a\prime}(f))({1}_{a})\\ & =({F}_{a,a\prime}(f)\circ {\alpha}_{a})({1}_{a})\\ & ={F}_{a,a\prime}({f}_{(}{\alpha}_{a}({1}_{a}))\\ & ={F}_{a,a\prime}(f)(x).\end{array}$$ \begin{aligned} ~~ alpha_{a'}(f)~~ \alpha_{a'}(f) &= \alpha_{a'}(\mathbf{y} a_{a,a'}(f)(1_a))\\ &= (\alpha_{a'} \circ \mathbf{y}a_{a,a'}(f))(1_a)\\ &= (F_{a,a'}(f) \circ \alpha_a)(1_a)\\ &= F_{a,a'}(f_(\alpha_a(1_a))\\ &= F_{a,a'}(f)(x). \end{aligned}

Thus, both composites are equal to identities. The proof of naturality follows from this. $\square$

The Yoneda embedding $\mathbf{y} : A \to \mathit{Set}^{A^{op}}$ is fully faithful.

**Proof.** By the Yoneda lemma, we have

$hom_{\mathit{Set}^{A^{op}}}(\mathbf{y}a,\mathbf{y}b) \cong \mathbf{y} b(a) \equiv hom_A(a,b)$

It is easy to check that this isomorphism is in fact the action of $\mathbf{y}$ on hom-sets. $\square$

category: category theory

Revision on September 19, 2018 at 14:11:49 by Ali Caglayan. See the history of this page for a list of all contributions to it.