Consider the usual model structure on $SSet$ (which is proper). Let $X$ and $Y$ be two fibrant simplicial sets and $f:X\rightarrow Y$ a fibration. If for any two vertices $x,x'$ of $X$, the homotopy pullback of $x$ and $x'$ is weakly equivalent to the homotopy pullback of $f(x)$ and $f(x')$. Can we deduce that the induced map on connected components $\pi_0(f):\pi_0(X)\rightarrow \pi_0(Y)$ is injective?
Actually I have proved that it induces isomorphisms on higher homotopy groups using the long exact sequence of homotopy groups. But on $\pi_0$ the injectivity seems not coming from the long exact sequence.
My stupid attempt: let $x$ and $x'$ be two vertices such that $f(x)$ and $f(x')$ are homotopic in $Y$. Then the homotopy pullback of $f(x)$ and $f(x')$ is weakly equivalent to the homotopy pullback of $f(x)$ and $f(x)$ (Cf. Hirschhorn Model categories and their localizations, Corollary 13.3.6). This implies the homotopy pullback of $x$ and $x'$ is weakly equivalent to the homotopy pullback of $x$ and $x$. But I don't know if the converse of Hirschhorn's corollary is true (and I think it is not true in general).
Edit: As the answer says, the converse is also true: Replace one of the points by a fibration and the homotopy pullback is the pullback. If the homotopy pullback is nonempty, then we have two points of the fibrant cofibrant replacement mapping to the two points of $X$, but the fibrant cofibrant replacement is weakly equivalent to a point.
If $x$ and $x'$ are in the same connected component, then as you say, their homotopy pullback is equivalent to the loop space based at $x$. If they're not in the same connected component, then their homotopy pullback is empty. So your claim is true. If you want to do things a bit more formally, you could replace the inclusions of $x$ and $x'$ with fibrations via path spaces to see the same result.