Proof

First of all, by “a diagram” in a derivator, we mean an object of $D(X)$ for some suitable category $X$. In the above case, $X$ is the category consisting of two commutative squares, as pictured above. We’ll write $abcdef$ for this $X$, and similarly $cdef$ for its lower-right L-shaped subcategory, and so on. We leave the verification of homotopy exactness of all squares to the reader.

Firstly, since the squares

are exact, if we start from a $cdef$-diagram and right Kan extend it to a full $abcdef$-diagram, then the right-hand square and outer rectangle must be pullback squares. Moreover, by composition of adjoints, right Kan extension from $cdef$ to $abcdef$ is equivalent to first extending to $bcdef$ and then to $abcdef$, and since the square

is exact, the left-hand square in such an extension must also be a pullback.

Now if we start with an $abcdef$-diagram, say $F$, we can restrict it to a $cdef$-diagram and then right Kan-extend it to a new $abcdef$-diagram. If $u\colon cdef \to abcdef$ is the inclusion, then this results in $u_\ast u^\ast F$, and we have a canonical natural transformation $\eta \colon F \to u_\ast u^\ast F$ (the unit of the adjunction $u^\ast \dashv u_\ast$). Since the square

is exact, the counit $u^\ast u_\ast G \to G$ is an isomorphism for any $G$, and in particular for $G=u^\ast F$, from which it follows by the triangle identities that $u^\ast F \to u^\ast u_\ast u^\ast F$ is also an isomorphism — i.e. the components of $\eta\colon F \to u_\ast u^\ast F$ at $c$, $d$, $e$, and $f$ are isomorphisms. Now if the right-hand square of $F$ is a pullback, then the restrictions of $F$ and $u_\ast u^\ast F$ to $bcef$ are both pullback squares; hence since the $cef$-components of $\eta$ are isomorphisms, so is the $b$-component. And if the left-hand square of $F$ is a pullback, then we can play the same game with $abde$ to show that the $a$-component of $\eta$ is an isomorphism, while if the outer rectangle is a pullback, we can play it with $acdf$. Hence in both of these cases, $\eta$ itself is an isomorphism, since all of its components are — and thus the remaining square in $F$ is also a pullback, since we have shown that it is so in $u_\ast u^\ast F$.

Proof

Since $f_! = j_! \bar{f}_!$ where $\bar{f}\colon K\to (J\setminus z)$ is induced by $f$ and $j\colon (J\setminus z) \to J$ is the inclusion, it suffices to suppose that $K = (J\setminus z)$. Now what we want is to prove that the following square is homotopy exact:

Exactness is trivial at all objects of $\square$ except $(1,1)$. In that case, we paste with another square:

The left-hand square is a comma square, hence homotopy exact, so it suffices to show that the composite square is homotopy exact. But the comma object associated to the cospan $\ast \to J \leftarrow (J\setminus z)$ is $(J \setminus z)/z$, and of course this comma square is also exact. And the composite square factors through this comma square by the functor $\Gamma \to (J \setminus z)/z$ which is assumed a nerve equivalence; hence it is also homotopy exact.