Image of a full functor is not necessarily a subcategory of the codomain I wonder wether I am right. If I am right, then the book Abstract and Concrete Categories is wrong. In Remark 4.2(1), the book says: "If $F: \mathbf{A} \to \mathbf{B}$ is a full functor or is injective on objects, then the image of $\mathbf{A}$ under $F$ is a subcategory of $\mathbf{B}$". In particular, this means that if $F: \mathbf{A} \to \mathbf{B}$ is a full functor, then the image of $\mathbf{A}$ under $F$ is a subcategory of $\mathbf{B}$. But I think I have found a counterxample for this. Consider, in $\mathbf{B}$, objects $B = F(A)$, $B' = F(A')$, $B'' = F(A'')$, and morphisms $g_0 = F (f_0): B \to B''$ and $g_1 = F(f' \circ f): B \to B''$, where $f: A \to A'$, $f': A' \to A''$ and $f_0: A \to A''$ are such that $f' \circ f \neq f_0$. If $g_0 = Ff' \circ Ff \neq g_1$, where the composition is in $\mathbf{B}$, then I think $F\mathbf{A}$ is not a subcategory of $\mathbf{B}$. Am I right?

I don't understand your argument, as pointed out in the comment : $F(f'\circ f) = F(f')\circ F(f)$ by definition of a functor.
However, you are correct that $FA$ is not always a subcategory (if $F$ is neither full nor injective on objects). Indeed, the formula $F(f')\circ F(f) = F(f'\circ f)$ does not hold if $\mathrm{dom}(f')\neq \mathrm{cod}(f)$, which can happen if $F$ is not injective on objects.
To give an explicit example, let $A$ have $4$ objects $a, b_0,b_1,c$ and two non identity arrows $a\to b_0, b_1\to c$.
Let $F$ be a functor such that $F(b_0) = F(b_1)$ and assume this is different from $F(a)\neq F(c)$. Then $F(b_1\to c)\circ F(a\to b_0)$ should be in $FA$ if it were a subcategory.
But this is a morphism $F(a)\to F(c)$ and so it should be a morphism $F(a\to c)$ for some $a\to c$. There is no such morphism in $A$ of course, so that $FA$ is not a subcategory.
However, if $F$ is full, something else saves us : $F(f)\circ F(g)$ is always in $FA$, as its domain is $F(\mathrm{dom}(g))$ and its codomain $F(\mathrm{codom}(f)))$, and $F$ is full, so there exists $h$ with $F(h) = F(f)\circ F(g)$ (although this need not be $f\circ g$, since the latter might not make sense).
So the statement is in fact correct.
