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Let $\mathscr{C}$ be a category of fibrant objects. For objects $X$, $Y$ of $\mathscr{C}$ we consider the category $\underline{\text{Hom}}(X,Y)$ of spans $X\leftarrow X'\rightarrow Y$, where $X'\to X$ is a trivial fibration. We can consider the full subcategory $\underline{\text{Hom}}'(X,Y)$, consisting of spans where the morphism $X'\to Y$ is a fibration.

I seem to have convinced myself, that $\underline{\text{Hom}}'(X,Y)\to\underline{\text{Hom}}(X,Y)$ induces a homotopy equivalence on geometric realizations. In particular, I can restrict to the subcategories $\underline{\text{Hom}}'(X,Y)$ to get the correct simplicial localization of $\mathscr{C}$.

Maybe someone knows that this is impossible? (The proof would go via constructing functorial fibrant resolutions for morphisms with target $Y$ using a fixed path object for $Y$.)

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Yes, one can use functorial factorisation (of weak equivalences with a fixed codomain) to show that two categories of spans are homotopy equivalent is correct.

Actually, the only subtle point I am aware of is in proving that your $\underline{\mathrm{Hom}} (X, Y)$ is equivalent to the space of hammocks from $X$ to $Y$ – for that one wants to use the homotopy calculus of right fractions introduced by Dwyer and Kan, so one has to show that the category of zigzags $$X \leftarrow \bullet \rightarrow \cdots \rightarrow \bullet \leftarrow Y$$ (with leftward-pointing arrows being weak equivalences) is weakly homotopy equivalent to the category of zigzags $$X \leftarrow \bullet \rightarrow \cdots \rightarrow Y$$ and the only way I know how to do this requires functorial factorisation of weak equivalences with a fixed domain! So if one has functorial path objects there is no problem at all.

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