求证0是复合映射g∘f的正则值——光滑流形证明困惑

First, let's recall what it means for 0 to be a regular value of (g \circ f): for every (x \in (g \circ f)^{-1}(0)), the differential (d_x(g \circ f): T_xX \to T_0\mathbb{R}^c) must be surjective.

Note that ((g \circ f)^{-1}(0) = f^{-1}(g{-1}(0)) = f^{-1}(Z)), so we only need to check the surjectivity condition for all (x \in f^{-1}(Z)). Let's walk through the key steps using the given conditions:

1. Chain Rule Setup
   By the chain rule, we know:
   [
   d_x(g \circ f) = d_{f(x)}g \circ d_x f
   ]
   Our goal is to show this composition is surjective onto (T_0\mathbb{R}^c).

2. Use Surjectivity of (d_{f(x)}g)
   The problem states that for all (y \in Y), (d_y g: T_yY \to T_{g(y)}\mathbb{R}^c) is surjective. For (x \in f^{-1}(Z)), (f(x) \in Z = g^{-1}(0)), so (d_{f(x)}g: T_{f(x)}Y \to T_0\mathbb{R}^c) is surjective. This means for any vector (v \in T_0\mathbb{R}^c), there exists some (u \in T_{f(x)}Y) such that (d_{f(x)}g(u) = v).

3. Decompose (u) Using the Given Sum Condition
   We're told that for all (x \in f^{-1}(Z)):
   [
   \text{im}(d_x f) + T_{f(x)}Z = T_{f(x)}Y
   ]
   But since (Z = g^{-1}(0)) and (g) has 0 as a regular value (because (d_y g) is surjective everywhere), the tangent space (T_{f(x)}Z) is exactly the kernel of (d_{f(x)}g) (a standard result for regular submanifolds: (T_y Z = \ker d_y g) when (y \in Z) and 0 is a regular value of (g)).

   So we can rewrite the sum condition as:
   [
   \text{im}(d_x f) + \ker(d_{f(x)}g) = T_{f(x)}Y
   ]
   This means our vector (u \in T_{f(x)}Y) (from step 2) can be written as:
   [
   u = d_x f(w) + k
   ]
   where (w \in T_xX) and (k \in \ker(d_{f(x)}g)).

4. Verify Surjectivity of the Composition
   Apply (d_{f(x)}g) to both sides of the decomposition:
   [
   d_{f(x)}g(u) = d_{f(x)}g(d_x f(w)) + d_{f(x)}g(k)
   ]
   Since (k \in \ker(d_{f(x)}g)), (d_{f(x)}g(k) = 0). And from step 2, (d_{f(x)}g(u) = v). So:
   [
   v = d_x(g \circ f)(w)
   ]
   This shows that for every (v \in T_0\mathbb{R}^c), there exists a (w \in T_xX) such that (d_x(g \circ f)(w) = v). In other words, (d_x(g \circ f)) is surjective.

Since this holds for all (x \in (g \circ f)^{-1}(0)), 0 is indeed a regular value of (g \circ f).

内容的提问来源于stack exchange，提问作者bphi