A.2.3 Lifting Criterion of Model Category, Lurie, Explanation The statement is Prop A.2.3.1, on page 811





I am confused with a few points:
(i) How is $C(A)\bigsqcup_A B \rightarrow C(B)$ a trivial cofibration.
(ii) Why can we regard $h$ as a homotopy from $g'$ to $g$
 A: Since $A$ is cofibrant each of the structure maps $A\rightarrow CA$ is a trivial fibration. Since the map $B\rightarrow CA\sqcup_AB$ is the pushout of one of these maps, it too is a trivial cofibration. Similarly the map $B\rightarrow CB$ is a trivial cofibration, and it follows immediately from the 2-of-3 property that the map 
$$k:CA\sqcup_AB\rightarrow CB$$
is a weak equivalence. 
To see that it is also a cofibration consider the following. There is a factorisation
$$k:CA\sqcup_AB\xrightarrow{k'} CA\sqcup_{A\sqcup A}(B\sqcup B)\xrightarrow{k''} CB$$
with the map $k''$ a cofibration by assumption, and so to see that $k$ is a cofibration it suffices to show that $k'$ is a cofibration. 
You can see that this is true by writing $k'$ as an induced map of pushouts and showing that it has the left lifting property with respect to acyclic fibrations. Setting down a putative lifting problem for $k'$ you get a lift over $CA$ for free. You get a lift over one factor in $B\sqcup B$ for free, and using the fact that $i:A\rightarrow B$ is a cofibration you construct a lift over the second summand which is compatible with the lift over $CA$. This gives you a lift over $CA\sqcup_{A\sqcup A}(B\sqcup B)$ as the induced map from the pushout.
It follows that $k'$, and hence $k$ is a cofibration. We have already seen that it is a weak equivalence.
For $ii)$ we get the claimed property of $h$ as follows. Since the map $j_0:B\rightarrow CB$ factors $B\rightarrow CA\sqcup_AB\rightarrow CB$ we get that one end of $h$ is $g'$. The other end of $h$ is some map whose claimed property we can identify using the following commutative diagram
$\require{AMScd}$
\begin{CD}
    A@>j_1>> CA\sqcup_ABE@>h_0>>X\\
    @ViV V   @VV  V@V=VV\\
    B @>j_1>> CB@>h>>X
\end{CD}
where the top left-hand horizontal map is the composition $A\xrightarrow{j_1} CA\rightarrow CA\sqcup_{A}B$ (note that the pushout $CA\sqcup_{A}B$ was formed from the span C$A\xleftarrow{j_0}A\xrightarrow{i}B$).
