Verifying theorem proofs derived from definition of homomorphism between structures I am following Hodges' Shorter Model Theory and I am trying to prove the following theorems (1.2.1 in the book), could anyone check my proof and see if they are ok please? In particular, I don't think I fully understand the clause about function preservation (2.3) and how to prove theorems about it.

If $A, B, C$ are $L-$structures and $f:A\to B$ and $g:B\to C$ are homomorphisms, then the composed map $gf$ is a homomorphism from $A$ to $C$. If moreover $f$ and $g$ are both embeddings then so is $gf$.

This means that I must prove that $gf: A\to C$ satisfies 2.1, 2.2 and 2.3 (see below).
2.1: For each constant $c$ of $L$, $g(f(c^A)=c^C$
$f:A\to B$ and $g:B\to C$ are homomorphisms, so we have $f(c^A)=c^B$ and $g(c^B)=c^C$. Substitute $c^B$ of the latter for $f(c^A)$ and we get $g(f(c^A)=c^C$.
2.2: For each $n>0$ and each $n$-ary relation symbol $R$ of $L$ and $n$-tuple $\bar{a}$ from $ R^A$, if $\bar{a}\in R^A$ then $gf\bar{a}\in R^C$.
Assume $\bar{a}\in R^A$, because $f$ is homomorphic, we get $f\bar{a}\in R^B$. But because $f\bar{a}$ is a tuple from B, and because $g$ is a homorphism, from $f\bar{a}\in R^B$ we get $gf\bar{a}\in R^C$.
2.3: For each $n>0$ and each $n$-ary function symbol $F$ of $L$ and $n$-tuple $\bar{a}$ from $ R^A$, $f(F^A(\bar{a}))=F^C(f\bar{a}))$. 
Not entirely sure if I formulated this 2.3 right for $gf$; what I can prove certainly doesn't match this statement.
Because $f$ is a homomorphism, $f(F^A(\bar{a}))=F^B(f\bar{a})$. But from what has been proved above (if $\forall \bar{a}:\bar{a}\in R^A$ then $f\bar{a}\in R^B$), and because $g$ being a homomorphism means $\forall \bar{b}: g(F^B(\bar{b}))=F^C(g\bar{b})$ (is it supposed to be $F$ or a new $G$ here?), we can substitute $\bar{b}$ with $f\bar{a}$ here.
i.e. $g(F^B(f\bar{a}))=F^C(g(f\bar{a})$ 

Let $A, B, C$ be $L$-structures. Then $1_A$ is an isomorphism. If $f:A\to B$ is an isomorphism then the inverse map $f^{-1}:dom(B)\to dom (A)$ exists and is an isomorphism from $B$ to $A$. If $f:A\to B$ and $g:B\to C$ are isomorphisms then so is $gf$.

$1_A$ is an isomorphism: 2.1 and 2.2 are trivial because the identity function always returns the same element you input. Thus, input $c^A$ you get $c^A$; likewise for 2.2. 
$f^{-1}$ is an isomorphism: by defintion of inverse function, if $f(c^A)=c^B$, then $f^{-1}(c^B)=c^A$. Likewise $f^{-1}(f(\bar{a}))\in R^A$. But I am not sure how to do it for the functions. For the embedding, obviously if $\bar{a}\in R^A \to f\bar{a}\in R^B$, then $f^{-1}f\bar{a}\in R^B \to \bar{a}\in R^A$. But I don't know how to prove that it is onto (in fact I don't even understand what it means to have an onto embedding).
$gf$ is an isomorphism: from the question above we already know $gf$ is a homomorphism, so we just need to prove that it is an isomorphism. i.e. $\bar{a}\in R^A \iff gf\bar{a}\in R^C$. Left to right follows from $gf$ being a homomorphism, so only need to prove right to left. Assume $gf\bar{a}\in R^C$, because $g$ and $f$ are isomorphic, $f\bar{a}\in R^B$ and then $\bar{a}\in R^A$. But then again I am stuck with the onto proof.

Definitions as noted in Hodges' book:

 A: *

*To show 2.3 holds for composition of two homomorphisms, I think what you've done is ok, but can be written more simply: $$\begin{align*}g \circ f(F^\mathcal{A}(\underline{a})) &= g F^\mathcal{B}(f\underline{a})
\\ & = F^\mathcal{C}(g\circ f \underline{a})
\end{align*}$$

*To show that 2.3 holds for the inverse $f^{-1}:B \to A$ of an isomorphic $f:A \to B$. $$\begin{align*}
f\left[f^{-1}(F^{\mathcal{B}}(\underline{b}))\right] & = F^{\mathcal{B}}(\underline{b})
\\ &= F^\mathcal{B}(f \circ f^{-1}\underline{b})
\\ & = f\left[F^\mathcal{B}(f^{-1}\underline{b})\right] \ \mbox{as $f$ is a homomorphism}
\end{align*}$$ $f$ is bijective, hence $f^{-1}(F^{\mathcal{B}}(\underline{b})) = F^\mathcal{B}(f^{-1}\underline{b})$. Moreover, $f^{-1}$ is simply bijective as the inverse of a bijective function. In particular, to see that it is onto:
$$a \in A \Rightarrow \exists b \in B: f(a)=b \Rightarrow f^{-1}(b)=a$$


*

*To show that a composition of two isomorphisms is an isomorphism, well you've already proved that the composition must be a homomorphism. Moreover, the composition of bijective functions must be bijective. In particular, to see that the composition is onto, suppose $f:A \to B$ and $G: B \to C$ are bijective (particularly onto), then for any $c \in C$, there is a $b \in B$ such that $g(b)=c$. Moreover, there is an $a \in A$ so that $f(a)=b$. Hence for any $c \in C$, there is an $a \in A$ so that $g \circ f(a)=c$.

