Transversality in the proof of the Blakers-Massey Theorem. Is it necessary? 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? 
 A: 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.  
A: I suggest you look at Mather's "Hurewicz Theorems for Pairs and Squares" in which Blakers-Massey is derived from his   Cube Theorems.  It works for all spaces, but the maps involved
have to be $m$- and $n$-connected with $m,n \geq 2$.
The basic inputs to the cube theorems are: (1)  Hurewicz/Dold local-to-global criteria for fibrations (or weak fibrations) and (2) the pullback of a cofibration by a fibration is a cofibration.
(Actually, (2) is not used in this proof.)
ADDING ON:  The proof is extremely cute and easy.  
Given a homotopy pushout square 
(spaces $A,B,C$ and $D$) pull back from the path fibration $\mathcal{P}(D) \to D$
to get another square.  The Second Cube Theorem tells you it is a homotopy pushout, 
and comparing connectivities of maps shows that it suffices to prove the B-M theorem
for the new square.
But the new square is a square of simply-connected spaces (since the spaces involved
are fibers of the maps involved), and the homotopy pushout is contractible.  Now the comparison map to the homotopy pullback may be identified (after suspension) with the inclusion of the wedge into the product, and that connectivity is easy to determine.
A: I refer here to my recent answer to 
What is the intuition behind the Freudenthal suspension theorem
where the results do not require simple connectivity for descriptions of the critical group, basically because the proofs do not use homological, i.e. abelian,  methods.  
May 30: The original Blakers-Massey results were related to triad homotopy groups, since the exact sequences involving these and relative homotopy groups showed the triad groups as the obstruction to excision. So there was a question of calculating these groups, and homology groups were used for this in the simply connected case, see the book by J.F. Adams A student's guide to algebraic topology. However such calculations in the non simply connected case do follow from a Generalised Van Kampen Theorem proved with J.-L. Loday. I have revised and updated a  paper of mine  ``Triadic Van Kampen theorems and Hurewicz theorems'', 
Algebraic  Topology, Proc. Int. Conf. Evanston March 1988, Edited
M.Mahowald and S.Priddy,  Cont. Math. 96 (1989) 39-57.
and made it available as 
http://pages.bangor.ac.uk/~mas010/pdffiles/VKTEVAN2.pdf
June 9, 2017  There is a nice book on "Cubical Homotopy" by Munson and Volic (CUP, 2015) which deals with a lot of these connectivity arguments. I am unclear whether these argument cover the connectivity results of Theorem 6.1 of
Brown, R. and Loday, J.-L.   ‘Homotopical excision, and Hurewicz theorems for n-cubes of spaces’. Proc. London Math. Soc. (3) 54 (1) (1987) 176–192,
which gives algebraic as well as connectivity results. 
