Pullbacks and fibers in the $\infty$-category of spaces Suppose given a commutative diagram in the $\infty$-category of spaces, as the one depicted below, where all but the bottom-right squares are pullbacks. Is it true that $H$ is (equivalent to) the fiber of the map induced between the fibers of $p$ and $q$ (or, equivalently, $e$ and $c$).
In particular, I would like to know whether it is true or not that $H$ is contractible iff the bottom-right square is a pullback.

Edit: I realized what I actually meant is that $H(a,b)$ (in place of $H$, and defined by such diagram) is contractible iff the map on homotopy fibers induced by $a$ (or, equivalently, by $b$) is a weak equivalence. But I guess the same counterexample should work.
 A: I think Denis Nardin's comment makes it clear that the answer should be no. To construct a counterexample, note that since $H = F \times_E F'$ and $F = fib(p)$, $F' = fib(e)$, the construction of $H$ depends only on $e$ and $p$. Whereas the total fiber (i.e. the fiber of the map between the fiber of $p$ and the fiber of $q$ or equivalently the fiber of the map between the fiber of $e$ and the fiber of $c$) depends in general on $q$ and $c$ as well. So we should be able to cook up a counterexample using the freedom to vary $q$ and $c$.
To make it very simple, let $e = p$ be the identity on a point, so that $q$ and $c$ are the inclusion of the basepoint of some space $C$. Then the total fiber is $\Omega^2 C$, wherease $H$ is a point. To complete the exercise, find a space $C$ such that $\Omega^2 C$ is not a point.
A: Well, I guess I can write as an answer what I wrote as a comment.
Any pullback square where $C$ is not discrete will yield a counterexample. For simplicity let $B=G=\ast$ and $C=S^1$. Then $E=H=\Omega S^1=\mathbb{Z}$ is not contractible but since the original square is a pullback, the induced map on the fibers of parallel arrows is necessarily a weak equivalence.
