Equivalence of categories of vector bundles and locally free sheaves Let $X$ be a scheme. It is known that the category $\mbox{Vec}_r(X)$ of vector bundles of rank $r$ on $X$ and the category $\mbox{Loc}_r(X)$ of locally free sheaves of rank $r$ on $X$ are equivalent (The equivalence is given by the functor $F\colon \mbox{Loc}_r(X)\rightarrow \mbox{Vec}_r(X)$ such that $F(\mathcal{E})=\textbf{Spec}(\mbox{Sym}(\check{\mathcal{E}}))$, where $\mbox{Sym}(\mathcal{F})$ is the symmetric algebra of $\mathcal{F}$ and $\textbf{Spec}(.)$ is as defined in Hartshorne p.128.
My confusion is the following. In many books it is written that a locally free subsheaf $\mathcal{E}'$ of a locally free sheaf $\mathcal{E}$ does not always correspond to a sub-vector bundle of the vector bundle $F(\mathcal{E})$. However, since the categories are equivalent, the monomorphism $\iota\colon\mathcal{E}'\rightarrow\mathcal{E}$ maps to a monomorphism $F(\iota)\colon F(\mathcal{E}')\rightarrow F(\mathcal{E})$ in $\mbox{Vec}_r(X)$. So, I would expect to get a sub-vector bundle of the vector bundle $F(\mathcal{E})$. 
A monomorphism in a category is not always an injective map in the usual sense. Is it maybe the point that $F(\iota)$ might not be an injective map, although it is a monomorphism?
PS: While I was trying to sort out this confusion, I looked in Hartshorne and was even more confused, because he does not define what a morphism of vector bundles is, although he defines what an isomorphism is (again on page 128). So my second question is what is the definition of a morphism of vector bundles in algebraic geometry?
 A: Let me tell the whole story of vector bundles because in many books this subject is not developed properly or only in an ad-hoc manner.
Definition. Let $X$ be a scheme. A pre-vector bundle on $X$ is an $X$-scheme $V$ such that for every $X$-scheme $T$ the set of $T$-valued points $V(T)=\hom_X(T,V)$ carries the structure of a module over $\mathcal{O}_T(T)$. This structure belongs to the data. For $X$-morphisms $T \to T'$ we require that $V(T') \to V(T)$ is linear over $\mathcal{O}_{T'}(T') \to \mathcal{O}_T(T)$. There is an obvious notion of a morphism of pre-vector bundles. Thus, we obtain a category. Every morphism $X' \to X$ induces a pullback functor $V \mapsto V|_{X'}$ (the underlying scheme is $V \times_X X'$) from pre-vector bundles on $X$ to pre-vector bundles on $X'$.
An example is $\mathbb{A}^n_X$ with the usual module structure on $\mathbb{A}^n_X(T) = \mathcal{O}_T(T)^n$. Any pre-vector bundle isomorphic to $\mathbb{A}^n_X$ for some $n$ is called trivial. A vector bundle over $X$ is a pre-vector $V$ bundle over $X$ which is locally trivial, i.e. there is an open covering $\{X_i \to X\}_i$ such that $V|_{X_i}$ is trivial for each $i$. We obtain a category of vector bundles.
If $E$ is a quasi-coherent module on $X$, then $\mathbb{V}(E):=\mathrm{Spec}(\mathrm{Sym}(\check{E}))$ is an affine $X$-scheme with $\mathbb{V}(E)(T) = \hom_{\mathcal{O}_T}(p^* \check{E},\mathcal{O}_T)$ for $p : T \to X$. This set has a canonical $\mathcal{O}_T(T)$-module structure, so that $\mathbb{V}(E)$ becomes a pre-vector bundle. Observe that $\mathbb{V}(\mathcal{O}_X^n)=\mathbb{A}^n_X$ is the trivial vector bundle. Hence, if $E$ is locally free, then $\mathbb{V}(E)$ is a vector bundle. Clearly this defines a functor $\mathbb{V}(-)$ from the category of locally free sheaves to the category of vector bundles over $X$.
Theorem. The functor $\mathbb{V}(-)$ is an equivalence of categories.
Proof. First, $\mathbb{V}(-)$ is fully faithful: Let $E,F$ be locally free sheaves on $X$. We claim that $\hom(E,F) \to \hom(\mathbb{V}(E),\mathbb{V}(F))$ is a bijection. Actually we have a morphism of sheaves $\underline{\hom}(E,F) \to \underline{\hom}(\mathbb{V}(E),\mathbb{V}(F))$ on $X$ and we claim that this is an isomorphism. This way we can work locally and thereby assume that $E,F$ are trivial, say $E=\mathcal{O}_X^n$ and $F=\mathcal{O}_X^m$.
Then $\underline{\hom}(E,F)$ becomes a sheaf of matrices $\mathcal{O}_X^{m \times n}$. The sheaf of $X$-morphisms $\mathbb{A}^n_X \to \mathbb{A}^m_X$ is $\mathcal{O}_X[t_1,\dotsc,t_n]^m$. One checks (!) that a morphism $\mathbb{A}^n_X \to \mathbb{A}^1_X$ corresponding to a polynomial in $\mathcal{O}_X[t_1,\dotsc,t_n]$ is a morphism of vector bundles iff this polynomial is linear. This shows $\underline{\hom}(\mathbb{V}(E),\mathbb{V}(F)) \cong (\mathcal{O}_X^m)^n$, which finishes the proof.
Secondly, $\mathbb{V}(-)$ is essentially surjective: Let $V$ be a vector bundle on $X$. Choose a cover $\{X_i \to X\}$ and a  trivialization $\phi_i : V|_{X_i} \cong \mathbb{V}(\mathcal{O}_{X_i}^n)$ (here $n$ may depend on $i$). By fully faithfulness, the resulting isomorphisms $\phi_{ij} := \phi_i |_{X_i \cap X_j} \phi_j^{-1}  |_{X_i \cap X_j}$ are induced by isomorphisms $\psi_{ij} : \mathcal{O}_{X_i \cap X_j}^n \cong \mathcal{O}_{X_i \cap X_j}^n$. Since $\phi_{ij}$ satisfy the cocycle condition, the same is true for the $\psi_{ij}$. Hence, we have a gluing data $(\mathcal{O}_{X_i}^n,\psi_{ij})$ and obtain a locally free sheaf $E$ such that $\mathbb{V}(E) \cong V$ by construction. $\square$
As already mentioned, every equivalence of categories preserves (and reflects) monomorphisms. In particular, a homomorphism of free sheaves of modules $\mathcal{O}_X^n \to \mathcal{O}_X^m$ is a monomorphism (in the category of locally free sheaves) if and only if the induced morphism of trivial vector bundles $\mathbb{A}^n_X \to \mathbb{A}^m_X$ is a monomorphism. However, monomorphisms of locally free sheaves should not be confused with monomorphisms in the larger category of all sheaves of modules, and monomorphisms of vector bundles should not be confused with monomorphisms of schemes, let alone morphisms of schemes whose underlying map of sets is injective, or injective on the fibers - which seems to be the usual ad-hoc definition of a subbundle.
For example, consider a regular global section $s$ of $\mathcal{O}_X$. Then $s : \mathcal{O}_X \to \mathcal{O}_X$ is a monomorphism of sheaves, hence also of locally free sheaves. The induced morphism of vector bundles $\mathbb{A}^1_X \to \mathbb{A}^1_X$ corresponds to the polynomial $s * t$. If $s$ is invertible, it is an isomorphism. If not, choose some $x \in X$ such that $s_x \in \mathfrak{m}_x$, i.e. $s(x)=0$ in $\kappa(x)$, so that on the fiber of $x$ we get the the zero morphism $\mathbb{A}^1_{\kappa(x)} \to \mathbb{A}^1_{\kappa(x)}$, whose underlying map of sets is not injective.
