Pullback bundle over a smooth manifold

Question

Following is something I found online.

More generally, if $f: M \longrightarrow N$ is a smooth map and $\pi: V \longrightarrow N$ is a vector bundle of rank $k$, there is a pullback bundle over $M$ : $$ f^{*} V \equiv M \times_{N} V \equiv\{(m, v) \in M \times V: f(m)=\pi(v)\} \stackrel{\pi_{1}}{\longrightarrow} M . $$ Note that $f^{*} V$ is the maximal subspace of $M \times V$ so that the diagram commutes. By the Implicit Function Theorem, $f^{*} V$ is a smooth submanifold of $M \times V$.

I have no idea what the Implicit function theorem is and I did some reading and it did not make any sense. I just have a feeling that there should be a way to give $f^*V$ a smooth structure without using this theorem but I just cannot figure it out. I would much appreciate it if someone can show me how this can be done. Thank you!

Answer 1

The twin results, the Inverse Function Theorem and the Implicit Function Theorem, are two of the most fundamental theorems in the study of smooth manifolds. If you're trying to understand the basics of smooth manifolds and submanifolds, you'll need to understand what these theorems say.

The Inverse Function Theorem

Given a smooth function $F \colon \mathbb{R}^n \to \mathbb{R}^n$, a natural question is whether an inverse function $F^{-1}$ exists. Equivalently, is $F$ bijective? Even when $n = 2$, this question is quite hard.

The Inverse Function Theorem does not quite answer this question, but it comes close. Namely, it provides an easily-checked criterion for an inverse function to exist locally.

Inverse Function Theorem: Let $F \colon \mathbb{R}^n \to \mathbb{R}^n$ be a smooth function. If the matrix of partial derivatives $DF(a)$ is invertible at some point $a \in \mathbb{R}^n$, then there are neighborhoods $U$ of $a$ and $V$ of $F(a)$ such that $F|_U \colon U \to V$ is a bijection (in fact, a diffeomorphism).

The Implicit Function Theorem

In the beginning, students learn two ways of describing curves in $\mathbb{R}^2$. One way is as a graph: the graph of $f \colon \mathbb{R} \to \mathbb{R}$ is the set $$\{(x,f(x)) \in \mathbb{R}^2 \colon x \in \mathbb{R}\}.$$ A second way is as a level set. The level set of $F \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ at $c \in \mathbb{R}$ is $$F^{-1}(c) = \{(x,y) \in \mathbb{R}^2 \colon F(x,y) = c\}.$$ Clearly, every graph is a level set: the graph of $f(x)$ is the level set $F(x,y) = y - f(x) = 0$. However, not every level set is a graph. For example, the circle $\{x^2 + y^2 = 1\}$ cannot be a graph of a function: it fails the "vertical line test."

A similar story goes through for surfaces in $\mathbb{R}^3$. One can talk about surfaces as graphs $z = f(x,y)$, or as level sets $F(x,y,z) = c$. Again, every graph is a level set, but the converse is generally false. This raises:

Question 1: Which level sets are graphs?

The Implicit Function Theorem does not quite answer Question 1, but it comes close. Namely, it provides an easily-checked criterion for a level set to be a graph locally.

In talking about level sets, we should worry about an even more basic concern. Namely: not all level sets are smooth (embedded) submanifolds. For example, the level set $\{(x,y) \in \mathbb{R}^2 \colon y^2 - x^3 = 0\}$ has a cusp at $(0,0)$. This raises:

Question 2: Which level sets are smooth embedded submanifolds?

A corollary to the Implicit Function Theorem will address this.

Implicit Function Theorem: Let $F \colon \mathbb{R}^n \times \mathbb{R}^k \to \mathbb{R}^k$ be a smooth function. Let $(x^1, \ldots, x^n, y^1, \ldots, y^k)$ denote the standard coordinates on $\mathbb{R}^n \times \mathbb{R}^k$. Write the $k \times (n+k)$-matrix of partial derivatives of $F$ at $(a,b) \in \mathbb{R}^n \times \mathbb{R}^k$ as $$DF(a,b) = \begin{pmatrix} \frac{\partial F^i}{\partial x^\ell}(a,b) & \frac{\partial F^i}{\partial y^j}(a,b) \end{pmatrix}.$$ Let $c = F(a,b)$. If the $k \times k$ submatrix given by $\begin{pmatrix} \frac{\partial F^i}{\partial y^j}(a,b) \end{pmatrix}$ is invertible, then there are neighborhoods $V \subset \mathbb{R}^n$ of $a$ and $W \subset \mathbb{R}^k$ of $b$, and a smooth function $f \colon V \to W$, such that the level set $F^{-1}(c) \cap (V \times W)$ restricted to $V \times W$ is exactly the graph of $f$.

Corollaries to the Implicit Function Theorem

The following result (addressing Question 2) is a corollary of the Implicit Function Theorem. It is used so often that texts often refer to it itself as the implicit function theorem.

Def: Let $F \colon M^m \to N^n$ be a smooth map between smooth manifolds. We say that $F$ is a submersion if for every $p \in M$, the derivative $DF(p) \colon \mathbb{R}^m \to \mathbb{R}^n$ is surjective.

Corollary A: If $F \colon M \to N$ is a submersion, then every level set $F^{-1}(c)$ is a smooth embedded submanifold of $M$.

This important fact can be generalized in several ways. Here's one such, which is relevant to your situation:

Def: We say that two smooth maps $F_1, F_2 \colon M \to N$ are transverse to each other if for every $x_1, x_2 \in M$ with $F_1(x_1) = F_2(x_2)$, we have that $DF_1(T_{x_1}M)$ and $DF_2(T_{x_2}M)$ together span the tangent space $T_{F_1(x_1)}N = T_{F_2(x_2)}N$.

Corollary B: If $F_1, F_2 \colon M \to N$ are transverse to each other, then $F_1^{-1}(F_2(M))$ is an embedded submanifold of $M$.

In your situation, we want to see the pullback bundle $$f^*V = \{(m,v) \in M \times V \colon (f \circ \pi_1)(m,v) = (\pi \circ \pi_2)(m,v)\}$$ as an embedded submanifold of $M \times V$. For this, check that $f \circ \pi_1$ and $\pi \circ \pi_2$ are transverse to each other, then apply Corollary B.