The Pullback Bundle is an Embedded Submanifold of its Parent Space

Question

$\newcommand{\mc}{\mathcal} \newcommand{\set}[1]{\{#1\}} \DeclareMathOperator{\pr}{pr} \newcommand{\at}{\big|} \DeclareMathOperator{\GL}{GL}$

Let $\pi:E\to N$ be a smooth vector bundle over a smooth manifold $N$ with typical fibre a $k$-dimensional vector space $\mc V$ and $f:M\to N$ be a smooth map between smooth manifolds.

Is it true that the total space of the pullback bundle $\Pi:f^*E\to M$ is an embedded submanifold of $M\times E$?

(For reference, I have described the pullback construction I below).

I think the above is true. It is clear that $f^*E$ has the subspace topology induced from $M\times E$. So we only need to check that the inclusion map is a smooth immersion.

When passing to charts, this is equivalent to checking that the map $(m, v)\mapsto (m, f(m), v):f^{-1}(U)\times \mc V\to f^{-1}(U)\times U\times \mc V$ is an immersion, where $U$ is an open set in $N$ over which $E$ trivializes. But this is indeed an immersion.

If the assertion is correct, then to check if a map $g:Q\to f^*E$ is smooth, we would just need to check if $\pr_1\circ g$ and $\pr_2\circ g$ are smooth (and this was the motivation for the question).

Thanks.

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Pullback Bundle Construction:

Let $\pi:E\to N$ be a smooth vector bundle over a smooth manifold $N$ with typical fibre a $k$-dimensional vector space $\mc V$ and $f:M\to N$ be a smooth map between smooth manifolds.

Define the set $f^*E=M\times_N E=\set{(p, x)\in M\times E:\ f(p)=\pi(x)}$. Define $\Pi:f^*E\to M$ as the restriction of $\pr_1:M\times E\to M$ to $M\times_N E$. We will write $(f^*E)_p$ to denote $\Pi^{-1}(p)$ for each $p\in M$.

Now for each smooth local trivialization $\phi:\pi^{-1}(U)\to U\times \mc V$ of $E$ over $U$, define a map $\Phi:\Pi^{-1}(f^{-1}(U))\to f^{-1}(U)\times \mc V$ as $\Phi(p, v)=(p, \pr_2\circ \phi(v))$ and note that $\Phi$ is a bijection whose restriction to $\Pi^{-1}(p)$ is a linear isomorphism from $(f^*E)_p$ to $\set{p}\times \mc V$ for each $p\in f^{-1}(U)$. For each $q\in U$, write $\phi_q$ to mean $\pr_2\circ \phi\at_{E_q}$ and note that $\phi_q:E_q\to \mc V$ is a linear isomorphism.

Let $\phi:\pi^{-1}(U)\to U\times \mc V$ and $\psi:\pi^{-1}(V)\to V\times \mc V$ be two overlapping smooth local trivializations of $U$ and $V$ over $V$ and $\Phi:\Pi^{-1}(f^{-1}(U))\to f^{-1}(U)\times \mc V$ and $\Phi:\Pi^{-1}(f^{-1}(V))\to f^{-1}(V)\times \mc V$ be the corresponding bijections. If $\tau:U\times V\to \GL(\mc V)$ is the transition function associated with $\psi\circ \phi^{-1}$, then note that $\Psi\circ \Phi^{-1}:f^{-1}(U\cap V)\times \mc V\to f^{-1}(U\cap V)\times \mc V$ is given by

\begin{equation*} \Psi\circ \Phi^{-1}(p, v)=(p, \tau_{f(p)}v) \end{equation*} for all $(p, v)\in (U\cap V)\times \mc V$. Defining $\Gamma:f^{-1}(U\cap V)\to \GL(\mc V)$ as $\Gamma=\tau\circ f$ we see that %equation \begin{equation*} \Psi\circ\Phi^{-1}(p, v) = (p, \Gamma_p(v)) \end{equation*} for all $(p, v)\in f^{-1}(U\cap V)\times \mc V$. It is clear that $\Gamma$ is smooth. Therefore there is a unique topology and smooth structure such that $\Pi:f^*E\to M$ is smooth vector bundle with typical fibre $\mc V$ whose smooth local trivializations are the maps $\Phi:f^{-1}(U)\to U\times \mc V$ as defined above.

Answer 1

$\newcommand{\M}{M}$ $\newcommand{\N}{N}$ $\newcommand{\brk}[1]{\left(#1\right)}$ $\newcommand{\be}{\beta}$ $\newcommand{\al}{\alpha}$ $\newcommand{\til}{\tilde}$

The pullback bundle is indeed an embedded submanifold of the product $M \times E$. The essential notion here is transversality. (Together with the fact that bundle projection is a submersion).

The full story with all the details is a bit long; I haven't seen it done in any textbook, I found all the steps here and there, and built my own picture of the things.

Definitions

$(1)$ Let $\M,\N$ be a smooth manifolds. Suppose $F:\N \to \M$ is a smooth map, $S \subseteq \M$ is an embedded submanifold. We say $F$ is transverse to $S$ if $\forall x \in F^{-1}\brk{S} \, , \, T_{F\brk{x}}\M= T_{F(x)}S + dF_x(T_x\N)$.

$(2)$ Let $\M,\N,\N'$ be smooth manifolds. Suppose $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. We say $F,F'$ are transverse to each other if $\forall x \in \N, x' \in \N' $ such that $F(x)=F'(x')$ , $T_{F(x)}\M=dF_x(T_x\N) + dF'_{x'}(T_{x'}\N')$.

Note: If either one of $F,F'$ is a submersion, then they are automatically transverse.

Note: Some of the proofs are at the end of the answer (So it will be possible to skim through the general scheme without all the details at first)

Lemma (1): $\M,\N$ be a smooth manifolds, $S \subseteq M$ is an embedded submanifold. Let $F:\N \to \M$ be transverse to $S$. Then $F^{-1}(S)$ is an embedded submanifold of $\N$ whose codimension is equal to the codimension of $S$ in $\M$.

proof: See Theorem 6.30 , in Lee. (pg 144).

Lemma (2): (This is exercise 13 in chapter 6, Lee)

Let $\M,\N,\N'$ be smooth manifolds. Suppose $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. Then $F,F'$ are transverse to each other if and only if the map $F \times F' : \N \times \N' \to \M \times \M$ is transverse to the diagonal $\Delta_\M = \{(x,x)|x \in \M \}$

Lemma (3): Let $\M$ be a smooth manifold. Then $\Delta_\M = \{(x,x)|x \in \M \}$ is an embedded (smooth) submanifold of $\M \times \M$.

Lemma (4):

Let $\M$ be a manifold. Let $\Delta_\M$ be the diagonal manifold of $\M$. (see Lemma 3 ). Then $T_{\brk{x,x}}\Delta_\M = \text{diag}\brk{T_x\M \times T_x\M}=\{(v,v)| v \in T_xM \}$. (i.e the tangent space of the diagonal is the diagonal of the tangent space).

proof: Since any tangent vector can be realized as a derivative of a path, the tangent space to a manifold is identical to the set of derivatives of paths. Since $\Delta_\M$ is an embedded submanifold, a path $\be:I \to \Delta_\M $ is smooth if and only if it is smooth when considered as a path into the product $\M \times \M$ if and only if each of its components is smooth. So $\be(t)=\brk{\al\brk{t},\al\brk{t}}$ , where $\al : I \to \M$, so $\dot \be (0) \overset{(*)}= \brk{\dot \al (0),\dot \al (0)}$ , hence its clear the tangent space to the diagonal is exactly the diagonal of the tangent space. (Where in (*) we used the canonical isomorphism between $T_{(x,x')}\brk{\M \times \M'} = T_x\M \oplus T_{x'}\M'$ via the differentials of the projections onto the different components).

corollary (1):

Let $\M,\N,\N'$ be smooth manifolds, $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. (In short we write $\N\overset{F}{\rightarrow}\M\overset{F'}{\leftarrow}\N'$). Assume $F,F'$ are transverse to each other. Then the fiber product of this diagram, which is defined as $\{\brk{x,x'} \in \brk{\N, \N'}|F(x)=F'(x') \}$ is an embedded smooth submanifold of the product $\N \times \N'$.

proof of corollary (1):

The fibered product $\N \times_{\M} \N'$ is the inverse image $(F \times F')^{-1}\brk{\Delta_{\M}}$. By Lemma 3, $\Delta_\M$ is a submanifold of $\M \times \M$. Now combine Lemma 2 and Lemma 1.

corollary (2):

Let $\M,\N,\N'$ be smooth manifolds, $\N\overset{F}{\rightarrow}\M\overset{F'}{\leftarrow}\N'$. If either one of $F,F'$ is a submersion, then the fiber product $\N \times_\M \N' = \{\brk{x,x'} \in \brk{\N, \N}|F(x)=F'(x') \}$ is an embedded submanifold of the product $\N \times \N'$.

proof of corollary (2): If one of $F,F'$ is a submersion, then these two maps are automatically transverse to each other. Now use corollary (1).

In particular we get the following proposition:

Let $\pi: E \to B$ be a vector bundle, $f:B' \to B$. The pullack bundle $f^*\brk{E}$ is an embedded submanifold of the product $B' \times E$. (This is becuse the bundle projection $\pi$ is always a submersion).

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proof of Lemma (2):

First, we need a sublemma: Sub-lemma:

Let $V$ be a vector space, $V_1,V_2 \subseteq V$ are subspaces. Let $\text{diag}(V \times V) = \{(v,v)|v \in V \} $. Then $V \oplus V = \text{diag}(V \times V) + \brk{V_1 \oplus V_2} \iff V = V_1 + V_2$

proof of the sublemma: $\Rightarrow :$ Let $v \in V$. Then $(v,0) \in V \oplus V$, hence by our assumption $\exists \til v \in V, v_1 \in V_1 , v_2 \in V_2$ such that $(v,0) = (\til v ,\til v) + (v_1,v_2)=(\til v + v_1, \til v + v_2) \Rightarrow \til v = -v_2, v = \til v + v_1 = v_1 - v_2 \in V_1 + V_2 $.

$\Leftarrow :$ Note that both sides of the left equation are subspaces. Hence, from symmetry it's enough to show that $\forall v \in V \, , \, (v,0) \in \text{diag}(V \times V) + \brk{V_1 \oplus V_2}$. The assumption $V =V_1 + V_2 \Rightarrow \exists v_i \in V_i$ such that $v = v_1 -v_2$. Define $\til v = -v_2$, so we get $(v,0)=(v_1-v_2,\til v +v_2)=(v_1 + \til v, v_2 + \til v) = (\til v,\til v) +(v_1,v_2)$.

Now to the actual proof of Lemma (2):

By definition (1), $F \times F'$ is transverse to the diagonal if

\begin{split} &\forall (x,x') \in (F \times F')^{-1}\brk{\Delta_\M} \, , \, T_{(F \times F')\brk{x,x'}}\brk{\M \times \M}= T_{(F \times F')(x,x')}\Delta_\M + d(F \times F')_{(x,x')}(T_{(x,x')}\brk{\N \times \N'}) \iff \\ & T_{\brk{F(x),F'(x')}}\brk{\M \times \M}= T_{\brk{(F(x),F'(x')}}\Delta_\M + d(F \times F')_{(x,x')}(T_x\N \oplus T_x\N') \iff \\ & T_{\brk{F(x),F(x)}}\brk{\M \times \M}= T_{\brk{(F(x),F(x)}}\Delta_\M + d(F \times F')_{(x,x')}(T_x\N \oplus T_x\N') \iff \\ &T_{\brk{F\brk{x}}}\M \oplus T_{\brk{F\brk{x}}}\M \overset{Lemma 4}= \text{diag}\brk{T_{F(x)} \M \times T_{F(x)}\M} + \brk{ dF_x \brk{T_x \N} \oplus dF'_{x'} \brk{T_{x'}\N'}} \overset{Sub-lemma} \iff \\ &T_{F(x)}\M =dF_x \brk{T_x \N} + dF'_{x'} \brk{T_{x'}\N'} \end{split}

Since the last row is the definition of transverse maps, we finished.

proof of Lemma (3): The diagonal is the graph of the smooth function $Id_\M$, and graphs of smooth functions are always embedded submanifolds of the product of the domain and the codomain. (See prop 5.4, Lee).