Exercises 3: Graph and Yoneda Lemma

EX 1: The Yoneda Lemma

Gr = 𝒢 ≝ \begin{xy} \xymatrix{ \ast \ar@/^/[d]^t \ar@/_/[d]_s \\ \ast } \end{xy}
\hat{𝒞} = Set^𝒞

1.

Y_0 = (\lbrace e \rbrace, ∅)

A graph homorphism from $Y_0$ to $G = (V, E)$ corresponds to picking a vertex in $G$, and conversely any such vertex defines a morphism from $Y_0$ to $G$.

One defines the graph

Y_1 ≝ v_1 \overset{f}{⟶} v_2

Similarly, a graph homorphism from $Y_1$ to $G = (V, E)$ corresponds to picking an edge in $G$, and conversely any edge defines a morphism from $Y_1$ to $G$.

2.

We want to construct a functor

Y: \begin{cases} 𝒢 ⟶ \hat{𝒢} = \text{Graphs} \\ 0 ⟼ Y_0 \\ 1 ⟼ Y_1 \end{cases}

but for any category $𝒞$, now.

The Yoneda functor:
Y: \begin{cases} 𝒞 ⟶ \hat{𝒞} ≝ Set^{𝒞^{op}} \\ A ⟼ 𝒞(\_, A) \end{cases}

3.

With 𝒞 = Gr ≝ \begin{xy} \xymatrix{ 0 \ar@/^/[d]^t \ar@/_/[d]_s \\ 1 } \end{xy}

If $G ∈ \hat{𝒢}$ is a graph with

  • $G_0$ as vertices
  • $G_1$ as edges
  • $G_s$ as source function
  • $G_t$ as target function

With the Yoneda functor:

  • $Y(0)$:

    • vertices: $Y(0)(0) = 𝒞(0, 0) = \lbrace id_0 \rbrace$
    • edges: $Y(0)(1) = 𝒞(1, 0) = ∅$
  • $Y(1)$:

    • vertices: $Y(1)(0) = 𝒞(0, 1) = \lbrace s, t \rbrace$
    • edges: $Y(1)(1) = 𝒞(1, 1) = \lbrace id_1 \rbrace$
  • source of $id_1$: $Y1s(id_1) = id_1 \circ s = s$

  • source of $id_1$: $Y1t(id_1) = id_1 \circ t = t$

4.

\begin{cases} \hat{𝒞}(YA, P) &≃ P(A) \\ θ &⟼ θ_A(id_A) \end{cases}

and

\begin{cases} P(A) &≃ \hat{𝒞}(YA, P) \\ x &⟼ θ \text{ defined by } θ_B(f) = Pf(x) \end{cases}

5.

$Y: 𝒞 ⟶ \hat{𝒞}$ is full and faithful: for all $A, B∈𝒞$,

\hat{𝒞}(YA, YB) ≃ YBA = 𝒞(A, B)

by the Yoenda lemma

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