As should be clear, I would like to know if it is true that a given commmutative square of spaces (i.e. simplicial sets) is a homotopy pullback iff the induced map on each homotopy fiber is a weak equivalence. More precisely, consider the following diagram:
$$\begin{array}{c}&&&&& A& \longrightarrow & B\\
&&&&&\downarrow && \downarrow \\
&&&&& C &\longrightarrow & D \\
&&&&\nearrow & &\nearrow\\
&&&1 & \longrightarrow & 1 \end{array}$$
Suppose that, after having taken homotopy pullbacks on both sides, the resulting front face is a homotopy pullback (i.e. the top horizontal arrow [between homotopy fibers] is a weak equivalence), and that this happens for any vertex of $C$. Is it true that the square involving $A,B,C,D$ is a homotopy pullback?


This seems to appear quite often but I couldn't find a precise reference with a proof.


Thanks in advance for your help.



Can homotopy pullbacks of spaces be checked on fibers?
$\require{AMScd}$I don't know a reference but the proof is easy enough. Form homotopy pullback squares


\begin{CD}
Fu @>>> Ff @>>> A \\
@VVV @VVV @V{u}VV \\
\* @>>> Fg @>>> P @>>> B \\
@. @VVV @VVV @VV{g}V \\
@. \* @>>> C @>>> D
\end{CD}


so that $Ff$, $Fg$ and $Fu$ are the homotopy fibers of $f$, $g$ and $u$ (where $f : A \to C$). By the assumption the map $Ff \to Fg$ is a weak equivalence and $Fu$ is its homotopy fiber so $Fu$ is (weakly) contractible. If you consider such diagrams for all points $\* \to C$, then you have tested homotopy fibers of $u$ at all points $\* \to P$. They are all contractible so $u$ is a weak equivalence and hence $A$ is the homotopy pullback of $g$ and $C \to D$ since $P$ is by construction.


