# coproduct of base and fiber is weakly equivalent to total space of a fibration in stable model category

Let $C$ be a proper pointed model category such that for any $X, Y \in C$ the natural morphisms $$QX \coprod QY \to X \coprod Y \to RX \times RY$$ are all weak equivalences (here $Q$ and $R$ are respectively the cofibrant and fibrant replacement functors). Also assume $HoC$ is additive (but I don't think it is relevant to my question). (For example, these conditions are satisfied when $C$ is a stable model category in the sense of Hovey chapter 7.)

Now let $f:E \to B$ be a fibration in $C$, with a section $s:B \to E$, and let $i:F \to E$ be the fiber of $f$. Why is the natural map $$i \coprod s:F \coprod B \to E$$ a weak equivalence? If it helps, you can assume $B$ is fibrant.

(This result is said to follow easily from the first paragraph above, according to p.26 Lemma 1.2.1.3 of Toen-Vezzosi Homotopical Algebraic Geometry II).

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well, i figured this out. the idea is to show a map $A \to B$ is a weak equivalence by showing $Hom(-, A) \to Hom(-, B)$ is a weak equivalence. See Hovey chapter 7. –  yogesh more Mar 19 '14 at 18:00