Quiver homomorphism

A homomorphism $𝑓 : Γ \to Δ$ of quivers is a pair of functions

𝑓 0 : Γ 0 \to Δ 0 𝑓 1 : Γ 1 \to Δ 1

on vertices and edges respectively such that #m/def/cat

𝑓 1 (Γ (𝑣, 𝑤)) \subseteq Δ (𝑓 0 (𝑣), 𝑓 0 (𝑤)) .

Regarding quivers as presheaves, this is precisely a natural transformation $Γ \to Δ$ , i.e. we have

(Δ 𝑠) (𝑓 1 (𝑎)) = 𝑓 0 ((Γ 𝑠) (𝑎)) (Δ 𝑡) (𝑓 1 (𝑎)) = 𝑓 0 ((Γ 𝑡) (𝑎))

Proof these are equivalent

Suppose ^M2 holds. Let $𝑏 \in 𝑓 1 (Γ (𝑣, 𝑤))$ , i.e. $𝑏 = 𝑓 1 (𝑎)$ for some $𝑎 \in Γ 1$ such that $(Γ 𝑠) (𝑎) = 𝑣$ and $(Δ 𝑠) (𝑎) = 𝑤$ , in which case

(Δ 𝑠) (𝑏) = 𝑓 0 ((Γ 𝑠) (𝑎)) = 𝑓 0 (𝑣) (Δ 𝑡) (𝑏) = 𝑓 0 ((Γ 𝑡) (𝑎)) = 𝑓 0 (𝑤)

and thus $𝑏 \in Δ (𝑓 0 (𝑣), 𝑓 0 (𝑤))$ . Therefore $𝑓 1 (Γ (𝑣, 𝑤)) \subseteq Δ (𝑓 0 (𝑣), 𝑓 0 (𝑤))$ .

Now suppose ^M1 holds. Let $𝑎 \in Γ 1$ , $𝑣 = (Γ 𝑠) (𝑎)$ , and $𝑤 = (Δ 𝑡) (𝑎)$ . Then $𝑎 \in Γ (𝑣, 𝑤)$ , so $𝑓 1 (𝑎) \in Δ (𝑓 0 (𝑣), 𝑓 0 (𝑤))$ , whence

(Δ 𝑠) (𝑓 1 (𝑎)) = 𝑓 0 (𝑣) = 𝑓 0 ((Γ 𝑠) (𝑎)) (Δ 𝑡) (𝑓 1 (𝑎)) = 𝑓 0 (𝑤) = 𝑓 0 ((Γ 𝑡) (𝑎))

as required.

These form the morphisms in $𝖰 𝗎 𝗂 𝗏$ .

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