Assume one is given a commutative square of spaces


$A \quad \to \quad C$


$
\downarrow \qquad \qquad \downarrow$


$B\quad \to \quad X$


which is a pushout and in which each map is a cofibration.
If $A \to B$ is $r$-connected and $A\to C$ is $s$-connected, 
then the Blakers-Massey theorem says that the square is 
$(r+s-1)$-cartesian (this means that the map from $A$ into the 
homotopy pullback of the remaining terms is $(r+s-1)$-connected).


The only proofs of the statement that I know of (at this level of
 generality) make use of transversality. However, if all spaces are  *simply connected,* there are
proofs which avoid transversality (for example, when $B$ is a contractible, one can 
deduce it using the Serre exact sequence).


**Question**:  *Is transversality intrinsic to a proof of the theorem in the general
 case?* 



Transversality in the proof of the Blakers-Massey Theorem. Is it necessary?
Take a look at the proof (attributed to Puppe) given in tom Dieck's new algebraic topology texbook (section 6.9). (I believe it also appears in tom Dieck, Kamps, Puppe (Lecture Notes in Mathematics 157).) This argument contains no obvious appeal to transversality. 


