为何自由R-模是投射模？基于自由模泛性质证提升性质

Let's start with quick refreshes to make sure we're on the same page:

• A free $R$-module over a ring $R$ is a module with a basis—meaning there’s a set of elements where every module element can be written uniquely as a finite $R$-linear combination of these basis elements.
• A projective $R$-module is defined by the lifting property: For any surjective $R$-homomorphism $ \pi: M \to N $ and any $R$-homomorphism $ f: P \to N $, there exists an $R$-homomorphism $ g: P \to M $ such that $ \pi \circ g = f $.

First: Why Free $R$-Modules Are Projective #

Take a free $R$-module $F$ with basis $ {e_i}_{i \in I} $. Suppose we have a surjective homomorphism $ \pi: M \to N $, and a homomorphism $ f: F \to N $. We need to build a lift $g: F \to M$ where $ \pi(g(e_i)) = f(e_i) $ for every basis element $e_i$.

Since $ \pi $ is surjective, for each $e_i$, there’s some $m_i \in M$ where $ \pi(m_i) = f(e_i) $. Using the free module’s universal property (we’ll dig into this next), we can define $g$ by setting $g(e_i) = m_i$ for each basis element, then extending linearly to all of $F$—this extension is unique and well-defined because every element of $F$ is a finite linear combination of the $e_i$.

Now confirm $ \pi \circ g = f $: For any basis element $e_i$, $ (\pi \circ g)(e_i) = \pi(m_i) = f(e_i) $. Since homomorphisms are fully determined by their action on a basis, this equality holds for all elements of $F$. That’s exactly the lifting property! So free modules satisfy the projective module definition.

Proving the Lifting Property Using the Free Module Universal Property #

First, let’s formalize the universal property clearly:

Universal Property of Free $R$-Modules: If $F$ is the free $R$-module on a set $S$ (so $S$ is a basis for $F$), then for any $R$-module $M$ and any function $ \phi: S \to M $, there exists a unique $R$-homomorphism $ \hat{\phi}: F \to M $ such that $ \hat{\phi}(s) = \phi(s) $ for all $s \in S$.

Here’s how we use this to prove the lifting property rigorously:

1. Let $F$ be free with basis $S = {e_i}_{i \in I}$. Let $ \pi: M \to N $ be surjective, and $ f: F \to N $ be a homomorphism.
2. For each $e_i \in S$, since $ \pi $ is surjective, pick some $m_i \in M$ where $ \pi(m_i) = f(e_i) $. This defines a function $ \phi: S \to M $ where $ \phi(e_i) = m_i $.
3. By the universal property, there’s a unique homomorphism $g: F \to M$ extending $ \phi $—so $g(e_i) = m_i $ for all $e_i$, and $g$ is linear over $R$.
4. Verify $ \pi \circ g = f $:
   • For any basis element $e_i$, $ (\pi \circ g)(e_i) = \pi(g(e_i)) = \pi(m_i) = f(e_i) $.
   • For any arbitrary element $ x = \sum_{i \in J} r_i e_i \in F $ (where $J$ is finite, $r_i \in R$):
     [
     (\pi \circ g)(x) = \pi\left(g\left(\sum r_i e_i\right)\right) = \pi\left(\sum r_i g(e_i)\right) = \sum r_i \pi(g(e_i)) = \sum r_i f(e_i) = f\left(\sum r_i e_i\right) = f(x)
     ]
   • Since this holds for all $x \in F$, $ \pi \circ g = f $.

The core idea here is that the free module’s basis lets us "pick preimages" for each basis element, and the universal property guarantees we can extend those picks to a full homomorphism that meets the lifting condition.

内容的提问来源于stack exchange，提问作者bateman