Let $(p,E,B)$ and $(p',E',B)$ be two $C^r$ vector bundles over the same base space $B$. (When $r>0$ all the spaces are $C^r$ manifolds and all the maps are $C^r$ smooth. When $r=0$ they are just topological. But I assume they are good enough: Hausdorff even paracompact or normal if you wish). Suppose $F:E \rightarrow E'$ is a monomorphism of vector bundles, i.e., $p= p'\circ F$ and $F_x:E_x \rightarrow E'_x$ is a linear injection.
My question is whether $F$ is an embedding ? I know that $F$ is injective and that it locally looks like an embedding.
We first show that $F(E)\subset E'$ as a subspace of $E'$ is a subbundle and hence a submanifold when $r>0$.
$\forall b\in B$, there exist trivialisations $(U,\phi)$ and $(U,\phi')$ of $(p,E,B)$ and $(p',E',B)$ with $b\in U$ such that the map $$U\times \mathbb{R}^{k} \cong p^{-1}(U) \xrightarrow{F} p'^{-1}(U) \cong U \times \mathbb{R}^n$$ is given by $$(x,v) \mapsto (x,F_x v)$$ where $F_x$ is an $n \times k$ matrix with rank $k$, and $x \mapsto F_x$ is a $C^r$ map. For simplicity we assume that $$F_b = \left(\begin{array} {c} M_b \\ * \end{array} \right)$$ where $M_b$ is a $k\times k$ matrix with $det(M)\neq 0$. By continuity of $x \mapsto F_x$, we assume that $$F_x = \left(\begin{array} {c} M_x \\ * \end{array} \right)$$ and $det(M_x) \neq 0, \forall x \in U$. Consider the map $$U \times \mathbb{R}^k \times \mathbb{R}^{n-k} \xrightarrow{G} U\times \mathbb{R}^n$$ $$(x,v,w) \mapsto (x,G_x(v,w)) = (x,F_xv+(0,w))$$ $G_x$ is a $n\times n$ matrix with $det(G_x) = det(M_x) \neq 0$ and $x \mapsto G_x$ is $C^r$. So $x \mapsto G_x^{-1}$ is well-defined and $C^r$ on $U$. Therefore $G$ is bijective $C^r$ and has the $C^r$ inverse $G^{-1}(x,z)=(x,G_x^{-1}z)$. Therefore $$G^{-1} \circ \phi':p'^{-1}(U) \cong U \times \mathbb{R}^k \times \mathbb{R}^{n-k}$$ is also a trivialisation of $(p',E',B)$. With this trivialisation we have $$U\times \mathbb{R}^{k} \cong p^{-1}(U) \xrightarrow{F} p'^{-1}(U) \cong U \times \mathbb{R}^k \times \mathbb{R}^{n-k}$$ $$(x,v) \mapsto (x,v,0)$$ so $p'^{-1}(U) \cap F(E) \cong U\times \mathbb{R}^{k} \times 0$ under the trivialisation $p'^{-1}(U) \cong U \times \mathbb{R}^k \times \mathbb{R}^{n-k}$. This proves that $F(E)$ as a subspace of $E'$ is a subbundle. And if $r>0$,$F(E)$ is also a submanifold of $E$. The map $p'^{-1}(U) \cap F(E) \cong U\times \mathbb{R}^{k} \times 0 \cong U\times \mathbb{R}^{k}$ gives a chart on $F(E)$.
Notice that $F$ is injective and that $\forall x \in E$, $T_xF$ is injective if $r>0$. We next prove that $F(E) \xrightarrow{F^{-1}} E$ is $C^r$, where $F(E)$ is given the subspace topology of $E'$. This will show that $F$ is an embedding.
Under the trivialisation above, locally we have $$U\times \mathbb{R}^{k} \cong U\times \mathbb{R}^{k} \times 0 \cong p'^{-1}(U) \cap F(E) \xrightarrow{F^{-1}} p^{-1}(U) \cong U \times \mathbb{R}^{k}$$ $$(x,v) \mapsto (x,v)$$ This map is $C^r$. So $F^{-1}$ is $C^r$ on $p'^{-1}(U) \cap F(E)$. Hence $F(E) \xrightarrow{F^{-1}} E$ is $C^r$.