0
$\begingroup$

Let $p_A:A\longrightarrow M$ and $p_B:B\longrightarrow M$ be two vector bundles. I know that a vector bundle morphism $\Phi:A\longrightarrow B$ is equivalent to a $C^\infty(M)$-linear map $\Phi_*:\Gamma(A)\longrightarrow \Gamma(B)$.

Now suppose $p_B:B\longrightarrow N$ is a vector bundle. Is it true that to give a vector bundle morphism $\Phi:A\longrightarrow B$ covering $\Phi_0:M\longrightarrow N$ is equivalent to give a $C^\infty(N)$-linear map $\Phi_*:\Gamma(A)\longrightarrow \Gamma(B)$ where the $C^\infty(N)$-module structre on $\Gamma(A)$ is induced by the morphism of algebras $\Phi_0^*:C^\infty(N)\longrightarrow C^\infty(M)$, $f\longmapsto f\circ \Phi_0$?

If not, what would be the analogous statement for this case?

Thanks.

$\endgroup$
0
$\begingroup$

I assume that $\Gamma(A)$ is the set of global sections of $A$. Consider $M$ be $\mathbb{R}^n, n>1$ and $N$ the point. Consider the trivial bundle $p_A=\mathbb{R}^n\times\mathbb{R}\rightarrow \mathbb{R}^n$ and $p_B:\mathbb{R}\rightarrow point$.

There exists a morphism of bundles $f:A\rightarrow B$ above the map $g:\mathbb{R}^n\rightarrow point$ defined by $f(x,y)=(point,y)$. But there does not a morphism $h:\Gamma(A)\rightarrow \gamma(B)$ above $g$. To see this, suppose that $h$ exists, and consider a section $s:\mathbb{R}^n\rightarrow A$ such that $s(x)=(x,u)$ $s(y)=(y,v), x,y\in \mathbb{R}^n, u\neq v$, we must have $f(s(x))=(point,u)=h(s)(point)=f(s(y))=(point,v)$ contradiction since $u\neq v$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.