# Abstract Nonsense

## Yoneda Lemma (Pt. II)

Point of Post: This is a continuation of this post.

Before we actually state Yoneda’s lemma we need to recall a technical detail we have often tacitly assumed about our categories. Namely, recall that a category is called locally small if the Hom sets are actually sets (opposed to the classes they are allowed to be). With this said we can now state and prove Yoneda’s lemma:

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Lemma (Yoneda): Let $\mathcal{C}$ be any locally small category, then for any object $X$ in $\mathcal{C}$ and any $F\in\mathcal{C}^\wedge$ one has an isomorphism

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$\text{Hom}_{\mathcal{C}^\wedge}(\mathfrak{Y}^\wedge(X),F)\cong F(X)\quad\mathbf{(1)}$

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Moreover, this isomorphism is natural in both $F$ and $X$.

Proof: To define the desired isomorphism (which is just going to be a bijection) $\eta_{X,F}:\text{Hom}_{\mathcal{C}^\wedge}(\mathfrak{Y}^\wedge(X),F)\to\ F(x)$ we note that given a natural transformation $\nu:\mathfrak{Y}^\wedge \implies F$ one has that $\nu_X$ is an arrow $\text{Hom}_\mathcal{C}(X,X)\to F(x)$, and so we define $\eta_{X,F}(\nu)=\nu_X(1_X)$. To show that $\eta_{X,F}$ really is a bijection we construct an inverse $\psi_{X,F}$. To do this, let $y\in F(X)$ be arbitrary, we want then to construct a natural transformation $\psi_{X,F}(y):\mathfrak{Y}^\wedge\implies F$. To do this we define $\psi(y)$ componentwise by $\psi_{X,F}(y)_z(k)=F(k)(y)$ where $z\xrightarrow{k}x$. Let’s show that this actually is a natural transformation. To this end, let $z\xrightarrow{f}z'$ be any morphism in $\mathcal{C}$ we then need to check that (recalling that our functors are contravariant)

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$F(f)\circ \psi_{X,F}(y)(z)=\psi_{X,F}(y)(z')\circ f^\ast$

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(where, as usual, $f^\ast$ is the image of $f$ under the functor $\text{Hom}(\bullet,X)$). So, for any $h\in\mathfrak{Y}^\wedge(X)(z)$ we check that

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\begin{aligned}\psi_{X,F}(y)(z'')(f^\ast(h)) &=\psi_{X,F}(y)(z'')(h\circ f)\\ &=F(h\circ f)(y)\\ &=F(f)(F(h)(y))\\ &=F(f)(\psi_{X,F}(y)(z')(h))\end{aligned}

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from where the naturality of $\psi_{X,F}(y)$ follows. Thus, we see that $\psi$ is really a mapping $F(X)\to\text{Hom}_{\mathcal{C}^\wedge}(\mathfrak{Y}^\wedge(X),F)$. Let’s now show that $\psi_{X,F}$ and $\eta_{X,F}$ are inverses of one another. Indeed, let $\nu$ be a natural transformation $\mathfrak{Y}^\wedge(X)\implies F$ we see then that for any $z\xrightarrow{h}X$ one has that

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$\psi_{X,F}(\eta_{X,F}(\nu))(h)=F(h)(\nu_X(1_X))=\nu_z\circ\mathfrak{Y}^\wedge(X)(y)(1_X)=\nu_z(h)$

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so that $\psi_{X,F}(\eta_{X,F}(\nu))=\nu$. Conversely, given any $y\in F(X)$ we have that

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$\eta_{X,F}(\psi_{X,F}(y))=F(1_X)(y)=y$

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so that $\eta_{X,F}$ and $\psi_{X,F}$ are indeed mutually inverses. Thus, we see that $\mathbf{(1)}$ indeed holds.

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The fact then that $\eta=\{\eta_{X,F}\}$ is really a natural transformation in either $X$ or $F$ if the other is fixed is just a simple computation, so I’ll leave it to you. $\blacksquare$

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While Yoneda’s lemma is itself interesting it’s real use is really to prove the following:

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Theorem: The Yoneda embedding $\mathfrak{Y}^\wedge:\mathcal{C}\to\mathcal{C}^\wedge$ really is an embedding.

Proof: The fact that $\mathfrak{Y}^\wedge$ is injective on objects is just the statement that we require Hom sets to be disjoint, the real crux of the argument is why it is full and faithful. What we really need to show is that $\mathfrak{Y}^\wedge$ induces bijections on the Hom sets.  In other words, we need to show that $\mathfrak{Y}^\wedge$ induces an isomorphism

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$\text{Hom}_\mathcal{C}(X,Y)\cong\text{Hom}_{\mathcal{C}^\wedge}(\mathfrak{Y}^\wedge(X),\mathfrak{Y}^\wedge(Y))$

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But, this is precisely Yoneda’s lemma as one can easily check. $\blacksquare$

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So now, the important consequence we’ve been referencing over and over again is

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Corollary: Let $X$ and $Y$ be objects in some locally small category $\mathcal{C}$. Then, if $\text{Hom}(\bullet,X)$ is naturally isomorphic to $\text{Hom}(\bullet,Y)$ then $X\cong Y$ by a unique isomorphism.

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Of course, the dual of the above statement proves that:

Corollary: Let $X$ and $Y$ be objects in some locally small category $\mathcal{C}$. Then, if $\text{Hom}(X,\bullet)$ is naturally isomorphic to $\text{Hom}(Y,\bullet)$ then $X\cong Y$ by a unique isomorphism.

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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 18, 2012 -

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