关于Mac Lane范畴论中首一均衡子同构性及红框语句的技术问询

Hey there, let's break these two category theory questions down clearly—they're key to wrapping your head around equalizers and isomorphisms!

1. Why is a monic equalizer an isomorphism? #

First, a quick foundational note: every equalizer is already a monomorphism by definition of its universal property. So when we talk about a "monic equalizer," we're really just referring to a standard equalizer—your question is probably getting at: when does that equalizer (which is always monic) turn out to be an isomorphism?

The short answer: When the two parallel morphisms it's equalizing are identical (i.e., f = g: A → B). Here's the step-by-step reasoning:

• Let e: E → A be the equalizer of f and g. By definition, f ∘ e = g ∘ e.
• If f = g, then the identity morphism id_A: A → A trivially satisfies f ∘ id_A = g ∘ id_A (both sides just equal f).
• By the universal property of equalizers, there must be a unique morphism d: A → E such that id_A = e ∘ d. That's our right inverse for e.
• Now check d ∘ e: E → E. We know f ∘ (e ∘ (d ∘ e)) = (f ∘ e) ∘ (d ∘ e) = (g ∘ e) ∘ (d ∘ e) = g ∘ (e ∘ d ∘ e) = g ∘ (id_A ∘ e) = g ∘ e = f ∘ e. Since e is an equalizer, the only morphism k: E → E making e ∘ k = e is id_E. So d ∘ e = id_E—that's our left inverse.
• With both a left and right inverse, e is an isomorphism.

Conversely, if e is an isomorphism, then f = f ∘ id_A = f ∘ (e ∘ e⁻¹) = (f ∘ e) ∘ e⁻¹ = (g ∘ e) ∘ e⁻¹ = g ∘ (e ∘ e⁻¹) = g ∘ id_A = g, so f must equal g.

2. Which part of the proof corresponds to the red-boxed statement in Mac Lane's Categories for the Working Mathematician? #

Since I can't see the exact red box, I'll go with the most common relevant passage in Mac Lane's book: the section stating that an equalizer is an isomorphism if and only if the parallel morphisms are equal.

The key part of the proof that links to that statement is two-fold:

1. Showing the identity morphism is an equalizer when f = g
   When f = g, id_A satisfies the equalizer condition (trivially, since f ∘ id_A = f = g = g ∘ id_A) and the universal property: any morphism h: X → A satisfies f ∘ h = g ∘ h, and it decomposes uniquely as h = id_A ∘ h.
2. Using the uniqueness of equalizers up to isomorphism
   Mac Lane emphasizes that any two equalizers of the same pair of morphisms are uniquely isomorphic. Since id_A is an equalizer (when f = g), the given equalizer e must be isomorphic to id_A—and any morphism isomorphic to the identity is itself an isomorphism.

If your red box is about a concrete example (like set theory, where the equalizer of f and g is the subset of A where f(x) = g(x)), the corresponding proof piece is pointing out that when f = g, this subset is all of A, so the inclusion map is the identity (a bijection, hence an isomorphism in Set).

内容的提问来源于stack exchange，提问作者Oliver G