# On vector bundle morphisms on different bases?

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.

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 $u\neq v$.