$0 \longrightarrow C(A)\otimes C(Y) + C(X)\otimes C(B) \longrightarrow C(X)\otimes C(Y) \longrightarrow C(X,A)\otimes C(Y,B) \longrightarrow 0$ In the relative version of Eilenberg-Zielber Theorem the following exact short sequence is taken
$$0 \longrightarrow C(A)\otimes C(Y) + C(X)\otimes C(B) \longrightarrow C(X)\otimes C(Y) \longrightarrow C(X,A)\otimes C(Y,B) \longrightarrow 0$$
Since given two complex $C,D$ we have  $(C \otimes D)_n := \bigoplus\limits_{p +q = n} C_p \otimes C_q$ I'm not confortable with the computation of exactness in the sequence above, i.e I don't understand why the sequence is exact.
Why $C(X)\otimes C(Y) / \left(C(A)\otimes C(Y) + C(X)\otimes C(B)\right) \simeq C(X,A)\otimes C(Y,B)$ ?
 A: I guess one important observation here is that both $C(X,A)$ and $C(Y,B)$ are free. Below is the proof:
First, we only need to focus on each degree of the sequence (since tensor product commutes with finite direct sum), i.e, we want to show that for each $p,q$, the sequence
$$0 \longrightarrow C(A)_p\otimes C(Y)_q + C(X)_p\otimes C(B)_q \longrightarrow C(X)_p\otimes C(Y)_q \longrightarrow C(X,A)_p\otimes C(Y,B)_q \longrightarrow 0$$
is exact.
Note that there each of the modules $ C(A)_p, C(Y)_q, C(X)_p, C(B)_q $ is free, so the inclusion map $C(A)_p \to C(X)_p$ induces the inclusion (since tensoring with a free module preserve inclusion):
$$C(A)_p \otimes C(Y)_q \to C(X)_p\otimes C(Y)_q$$
Similarly, we have an inclusion map $C(X)_p\otimes C(B)_q \to C(X)_p\otimes C(Y)_q$, hence, we have the exact sequence
$$0 \to C(A)_p\otimes C(Y)_q + C(X)_p\otimes C(B)_q \to C(X)_p\otimes C(Y)_q$$
On the other hand, since tensor product preserves surjective map (right exact), we have the surjections
$$C(X)_p\otimes C(Y)_q \to C(X)_p\otimes C(Y,B)_q \ \text{   and   }\  C(X)_p\otimes C(Y,B)_q \to C(X,A)_p\otimes C(Y,B)_q \to 0$$
which implies that we have a surjection
$$C(X)_p\otimes C(Y)_q \longrightarrow C(X,A)_p\otimes C(Y,B)_q \longrightarrow 0$$
Lastly, we want to show that
$$C(A)_p\otimes C(Y)_q + C(X)_p\otimes C(B)_q \stackrel{f}{\to} C(X)_p\otimes C(Y)_q \stackrel{g}{\to} C(X,A)_p\otimes C(Y,B)_q$$
is exact.
One implication is simple, take any $a\otimes y + x\otimes b \in C(A)_p\otimes C(Y)_q + C(X)_p\otimes C(B)_q$ then
$$g\circ f (a\otimes y + x\otimes b) = g(a\otimes y + x\otimes b) = g(a\otimes y) +g(x\otimes b) = 0$$
(where the last equality holds since, for example $g(a\otimes y)=(a+C(A)_p)\otimes (y+ C(B)_q)=0\otimes (y+ C(B)_q) =0$), showing that $\text{Im}(f) \subseteq \text{Ker}(g)$.
Conversely, suppose that $c=\sum X_i\otimes Y_j \in \text{Ker}(g) \setminus \text{Im}(f)$, since the sum is a finite sum and the modules $C(X)_p,C(Y)_q$ are free, we can assume that there are finite basis elements $x_i \in C(X)_p \setminus C(A)_p$ and $y_j\in C(Y)_q\setminus C(B)_q$ such that
$$c=\sum c_{ij}x_i\otimes y_j$$
Note that the elements $x_i + C(A)_p$ and $y_j+C(B)_q$ are basis elements of $C(X,A)_p$ and $C(Y,B)_q$ respectively. Therefore, the fact that $g(\sum c_{ij}x_i\otimes y_j) =0$ implies that $c_{ij}=0$ for all $i,j$, hence $c=0$.
