关于将简单图边替换为余张的函子sGrph→Quiv是否为左伴随的问询

Great question! Let's break this down clearly, starting with formalizing the setup and then using adjoint functor theory to confirm the answer.

First, let's align on the core definitions:

• sGrph: The category of simple undirected graphs (no loops, no multiple edges), where morphisms are graph homomorphisms (vertex maps that preserve edge adjacency).
• Quiv: The category of directed graphs (quivers, which allow loops and multiple edges), where morphisms are directed graph homomorphisms (vertex maps that preserve directed edges).
• Your functor F: sGrph → Quiv: For any simple graph G, F(G) shares the same vertex set as G; each undirected edge {u,v} ∈ E(G) is replaced by a cospan. Since you specified vertices stay unchanged, this means we're adding both directed edges u → v and v → u to F(G) for each undirected edge in G.

You already observed F preserves colimits—this is a huge hint, because left adjoints always preserve colimits. To confirm F is a left adjoint, we just need to construct its right adjoint (or verify the Special Adjoint Functor Theorem conditions, but constructing the adjoint is more concrete).

Step 1: Define the Right Adjoint Functor #

Let's build a functor G: Quiv → sGrph that acts as F's right adjoint:

• For a quiver Q, G(Q) has the same vertex set as Q. An undirected edge {u,v} exists in G(Q) if and only if Q contains both directed edges u → v and v → u.
• For a directed graph homomorphism f: Q₁ → Q₂, G(f) uses the same vertex map as f. Since f preserves directed edges, if {u,v} is an edge in G(Q₁), Q₂ must have f(u) → f(v) and f(v) → f(u)—so {f(u), f(v)} is an edge in G(Q₂), making G(f) a valid simple graph homomorphism.

Step 2: Verify the Adjoint Isomorphism #

We need a natural bijection between these morphism sets for any simple graph G and quiver Q:

Hom_sGrph(G, G(Q)) ≅ Hom_Quiv(F(G), Q)

Here's why this holds:

1. From sGrph to Quiv: Take a simple graph homomorphism h: G → G(Q). For every undirected edge {u,v} ∈ E(G), {h(u), h(v)} is an edge in G(Q)—meaning Q has both h(u) → h(v) and h(v) → h(u). We can turn h into a directed homomorphism F(h): F(G) → Q using the same vertex map; since F(G) has u → v and v → u for each undirected edge in G, F(h) preserves all directed edges of F(G).
2. From Quiv to sGrph: Take a directed homomorphism k: F(G) → Q. For every undirected edge {u,v} ∈ E(G), F(G) has u → v and v → u, so Q must have k(u) → k(v) and k(v) → k(u)—this means {k(u), k(v)} is an edge in G(Q), so k (as a vertex map) is a valid simple graph homomorphism G → G(Q).

This bijection is natural in both G and Q, so F is left adjoint to G.

Bonus: SAFT Confirmation #

Since both categories are locally small, Quiv is complete (has all limits), and F preserves colimits, the Special Adjoint Functor Theorem (SAFT) tells us F has a right adjoint if it satisfies the solution set condition. Our explicit construction of G proves this condition holds, so this is another way to confirm F is a left adjoint.

内容的提问来源于stack exchange，提问作者Joe Moeller