How to show $f^*E$ is a smooth submanifold...

Question

I'm wondering how to show the following: let $E$, $B_1$ and $B_2$ smooth manifolds. Suppose $\rho:E\rightarrow B_2$ is a smooth vector bundle and $f:B_1\rightarrow B_2$ a smooth map. If we write $$f^*E:=\{(b, e)\in B_1\times E: \rho(e)=f(b)\},$$ how can I show $f^*E$ is a smooth submanifold of $B_1\times E$. Any help will be valuable, thanks...

Answer 1

So you'll probably want to check next that it's a vector bundle. But, first, choose local coordinates on $B_1$, $B_2$, and $E$ (for the last, preferably taking the local product structure into account). If $\dim B_1=m$, $\dim B_2=n$, and $\dim E=n+k$, working in local coordinates, $f^*E$ is defined by the equation $F(x,y)=\tilde f(x)-\tilde p(y)=0$, where $F$ maps (an open set in) $\Bbb R^{m+n+k}$ to (an open set in) $\Bbb R^n$. Because $p$ is a submersion, so is $F$, and so $0$ is a regular value.

Answer 2

First if $U\subseteq B_1$ is open then $f^*E\cap (B_1\times E_{|U})=f_{|U}^*E_{|U}$, so since being a smooth submanifold is a local property, it is enough to find an open covering $\mathcal U$ of $B_1$ such that for each $U\in\mathcal U$ $f_{|U}^*E_{|U}$ is a smooth submanifold of $B_1\times E_{|U}$. Therefore it is enough to show the statement for trivial bundles.

If $E$ is trivial, there is a finite dimensional vector space $V$ and a bundle isomorphism $\Phi: E\to B_2\times V$. Then the diffeomorphism $\mathrm{id}_{B_1}\times\Phi$ maps $f^*E$ onto $f^*(B_2\times V$) so we can assume $E=B_2\times V$.

The map $$g:B_1\times V\to B_1\times (B_2\times V),\; (b,v)\mapsto (b,(f(b),v))$$

satisfies $\mathrm{im}(g)=f^*E$ and has a smooth retraction $h:B_1\times (B_2\times V)\to B_1\times V$ with $h\circ g=\mathrm{id}$. From $h\circ g=\mathrm{id}$ it follows that $g$ is a smooth immersion which is also a homeomorphism onto its image. Therefore $\mathrm{im}(g)=f^*E$ is a smooth submanifold of $B_1\times E$.