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Presheaves, Yoneda embedding, and Yoneda lemma [math-001A]

Definition. Presheaf [local-0]

Let AA be a category. A presheaf over AA is a functor of the form

X:AopSetX : A^{op} \to Set

For each object aAa \in A, we will denote by

Xa:=X(a)SetX_a := X(a) \in Set

the evaluation of XX at aa. The set XaX_a will sometimes be called the fibre of the presheaf XX at aa, and the elements of XaX_a thus deserve the name of sections of XX over aa.

For morphism auba \xrightarrow{u} b, the induced map from XbXaX_b \to X_a denotes u:=X(u)u^* := X(u).

The category of presheaves over AA denotes A^\widehat{A}.

Definition. Yoneda embedding [local-1]

Yoneda embedding is a functor

h:AA^h(a):=HomA(,a)\begin{aligned} &h &&: A \to \widehat{A} \\ &h(a) &&:= \text{Hom}_A(-,a) \end{aligned}

we denote ha:=h(a)h_a := h(a)

Lemma. Yoneda [local-3]

For any presheaf XX over AA, there is a natural bijection of the form

θ:HomA^(ha,X)Xaθ(α):=αa(1a)\begin{aligned} &\theta &&: \text{Hom}_{\widehat{A}}(h_a,X) \xrightarrow{\sim} X_a \\ &\theta(\alpha) &&:= \alpha_a(1_a) \end{aligned}

Proof. [local-2]

We first define inverse map τ:XaHomA^(ha,X)\tau : X_a \to \text{Hom}_{\widehat{A}}(h_a,X), given a section ss of XX over aa, i.e. sXas \in X_a, we have

τ(s)b:HomA(b,a)Xbτ(s)b(f):=f(s)\begin{aligned} \tau(s)_b &: \text{Hom}_A(b,a) \to X_b \\ \tau(s)_b(f) &:= f^*(s) \end{aligned}

for each morphism bfab \xrightarrow{f} a. This indeed defines a morphism haτ(s)Xh_a \xrightarrow{\tau(s)} X in A^\widehat{A}.

Now check θ\theta and τ\tau indeed are inverse of each other. First: given sXas \in X_a, we have

θ(τ(s))=τ(s)a(1a)=1a(s)=X(1a)(s)=s\theta(\tau(s)) = \tau(s)_a(1_a) = 1_a^*(s) = X(1_a)(s) = s

Another direction: given α:haX\alpha : h_a \to X, we have

τ(θ(α))b(f)=τ(αa(1a))b(f)=f(αa(1a))by definition of τ=X(f)(αa(1a))=αb(HomA(f,a)(1a))by naturality=αb(1af)=αb(f)\begin{aligned} \tau(\theta(\alpha))_b(f) &= \tau(\alpha_a(1_a))_b(f) \\ &= f^*(\alpha_a(1_a)) \quad \text{by definition of } \tau \\ &= X(f)(\alpha_a(1_a)) \\ &= \alpha_b(\text{Hom}_A(f, a)(1_a)) \quad \text{by naturality} \\ &= \alpha_b(1_a \circ f) \\ &= \alpha_b(f) \end{aligned}

for each f:baf : b \to a. Naturality is obvious, hence they form a natural bijection.