# Homotopy pullback square implies weak equivalence of homotopy fibres

I am quite confused about the following situation: suppose that we have a map $f \colon X \to Y$. Its homotopy fibre is defined as the pullback of the following of diagram:

\begin{matrix} Ff & {\rightarrow} & * \\ \downarrow{} && \downarrow{} \\ X & \rightarrow{} & Y \end{matrix}

Now if $g \colon W \to Z$ is another map and the diagram \begin{matrix} W & {\rightarrow} & X \\ \downarrow{} && \downarrow{} \\ Z & {\rightarrow} & Y \end{matrix}

is homotopy cartesian, why does it follow that the homotopy fibres of $f$ and $g$ are weakly equivalent?

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In ordinary category theory, if I delete the word "homotopy" and replace "weakly equivalent" by "isomorphic", this follows from the pullback pasting lemma. I imagine something similar is true in the homotopy world. –  Zhen Lin Dec 2 '12 at 14:33