(I posted this question on Mathstack but I haven't received any answers or comments so I thought I might as well try my luck here. I apologize if it is not an appropriate question.)


Theorem (Quillen) Let $F:\mathcal{C} \rightarrow \mathcal{D}$ suppose that $ N(F\downarrow d)$ is weakly equivalent to $\*$ for all $d \in \mathcal{D}$ then $NF: N\mathcal{C} \rightarrow N\mathcal{D}$ is a weak equivalence. 


The proof goes as follows: First we prove that if $F :\mathcal{C} \rightarrow \mathcal{D}$ is a functor, it is homotopy terminal if and only if $N(d \downarrow F)^{op}$ is weakly equivalent to a point for all $d \in \mathcal{D}$. If I can show/understand why $N(F\downarrow d)$ weakly equivalent to $ \*$ implies that $N(d \downarrow F)^{op}$ is weakly equivalent to $\*$ then the theorem follows easily ( by taking $X = Const\_\mathcal{D} \*$ in the definition of homotopy terminal ). So my question is:


Why does $N(F\downarrow d)$ weakly equivalent to $ \*$ for all $d \in \mathcal{D}$ imply that $N(d \downarrow F)^{op}$ is weakly equivalent to $\*$ for all $d \in \mathcal{D}$? 



A question about the proof of Quillen's Theorem A
If $S \colon \mathcal A \to \mathcal B$ is a functor, let $\newcommand{\op}{\mathrm{op}}S^\op \colon \mathcal A^\op \to \mathcal B^\op$ be its opposite. Then $(S \downarrow T)^\op \cong T^\op \downarrow S^\op$. Combine this with the equality $NF = NF^\op$ (if you identify $N\mathcal C = N\mathcal C^\op$, etc.)


