Let $M$ be a very nice model category (cofibrantly generated, combinatorial or cellular and left proper simplicial model category). Let $f: X\rightarrow Y$ and $g: X\rightarrow Z $ be two morphisms in $M$ such that $g$ is weak equivalence. Suppose that the map $r: Z\rightarrow Y\cup\_{X} Z $ is a weak equivalence in the localized model category $\mathrm{L}\_{\{ f\}}M$. Is it true that $\mathrm{L}\_{\{ f\}}M=\mathrm{L}\_{\{ r\}}M$ ?



Localization of a model category
No. Let $M$ be the category of simplicial sets with the Kan model structure. Let $S^1$ be $\Delta^1$ with its endpoints identified and let $f : \Delta^1 \to S^1$ be the obvious map. Let $g : \Delta^1 \to \Delta^0$ be the unique map. Then $r : \Delta^0 \to \Delta^0$ is the identity map, so it is a weak equivalence even in $M$, so $L\_{\{r\}}M = M$, but $L\_{\{f\}}M \neq M$ since $S^1$ is not equivalent to $\Delta^1$ in $M$.


