I am reading the book Triangulated Categories by Neeman. I have come across a sentence and I'm not really sure what it is trying to say. For those with access to the book, it is Remark 2.1.23.
Let $\mathcal{D}$ be a triangulated category with triangulated subcategory $\mathcal{C}$. Denote by $\text{Mor}_{\mathcal{C}}$ the class of morphisms whose mapping cone is in $\mathcal{C}$. In the usual way we can form the Verdier localisation which I will denote by $\mathcal{D}/\mathcal{C}$ with a quotient functor $Q : \mathcal{D} \rightarrow \mathcal{D}/\mathcal{C}$. The morphisms are given by roofs (or spans) and the composite of two composable spans is given by taking the homotopy pullback of their summits.
A composite of spans then looks something like this (diagram taken from the text) http://tinypic.com/r/2up360i/9
where $P$ is the homotopy pullback of $f_{2}$ and $g_{1}$. Here $v'$, $f_{1}$, and $f_{2}$ are in $\text{Mor}_{\mathcal{C}}$, and $$Q(g_{2})Q(f_{2})^{-1} Q(g_{1})Q(f_{1})^{-1} = Q(g_{1})Q(u')Q(v')^{-1}Q(f_{1})^{-1}. $$ The text then remarks that,
In other words, to give two composable morphisms $X \rightarrow Y$ and $Y \rightarrow Z$ in $\mathcal{D}/\mathcal{C}$, and their composite, is nothing other than to give $X' \rightarrow Y'$, and $Y' \rightarrow Z'$ in $\mathcal{D}$ and maps $X' \rightarrow X$, $Y' \rightarrow Y$, and $Z' \rightarrow Z$ in $\text{Mor}_{\mathcal{C}}$ expressing the isomorphism of hte composable pair in $\mathcal{D}/\mathcal{C}$ with a composable pair in $\mathcal{D}$.
I have no idea what is meant by this remark. How does the argument to that point imply that? I am struggling to see why it is true, and struggling even more to see how it is simply a rephrasing of the equality above. How is a composite of morphisms in the quotient category determined that way, and moreover, what precisely is meant by the "expressing the isomorphism of the composable pair..."?
Any help is appreciated.
Comparing Neeman's notation to the picture, $X',Y',Z'$ become $ P, W_2, Z$ while $X,Y,Z$ are the same in both places. The vertical morphisms in the picture are the isos Neeman describes; in the picture, the isomorphism $Z'\to Z$ is trivial, but of course it could be nontrivial and we'd still have a well defined morphism. In particular $g_2$ is determined as a map in $D/C$ by the other data, so superfluous.