Pullbacks and homotopy equivalences Say I have a map between pullback squares $(Y \rightarrow Z \leftarrow X) \to (Y' \rightarrow Z' \leftarrow X')$.  If the maps $X \to X'$, $Y \to Y'$ and $Z \to Z'$ are homotopy equivalences, does it follow that the induced map $X \times_Z Y \to X' \times_{Z'} Y'$ between the pullbacks is also a homotopy equivalence?  If not, what additional conditions (e.g., insisting that $X \to Z$ is a fibration, $Y' \to Z'$ is a cofibration, everything is a CW complex, etc.) are needed?  
I'll also like to know the answer in the case of pushout squares, but I guess I can just dualize whatever the answer to the previous question turns out to be.  
I tried to construct an inverse map directly using the homotopy inverses $X' \to X$, $Y' \to Y$, and $Z' \to Z$, but I could not guarantee that the resulting diagram commutes enough to produce a map $X' \times_{Z'} Y' \to X \times_Z Y$.  Even then, I'm not certain that I can somehow glue the homotopies in a compatible way to prove that the constructed map is a homotopy inverse.  

I'll answer the question in the pushout case.
If one of the maps of each pushout square is a cofibration, the induced map will be a homotopy equivalence, see Proposition 5.3.4 of Tammo tom Dieck's "Algebraic Topology". You don't need any further conditions on the spaces for this, not even that they are compactly generated or of the homotopy type of a CW complex.
The reason for this being true is that the cofibration-condition guarantees that both pushouts will be homotopy pushouts.
As you guessed, the dual statement for pullbacks is also true.
