Hungerford Algebra - Chapter IV - proof of the splitting lemma

Im working with the splitting lemma right now and I don't understand it too much. If someone has that book and could help me complete the sketch in page 177, proof of 1.18) I would be very grateful.

the book can be found here also http://books.google.se/books?id=t6N_tOQhafoC&printsec=frontcover&hl=sv&source=gbs_atb#v=onepage&q&f=false

-
The proof is long though not particularly hard: what part(s) exactly are being difficult to you? Why don't you write down in your question what have you done and where precisely you don't understand? –  DonAntonio Nov 21 '12 at 11:19
I dont know what they mean with "the diagram commutes" which means I dont really know how to use gf=0 and gh=id(A2). –  grendizer Nov 21 '12 at 11:44
A diagram with a mesh with vertices $A,B,C,D$ surrounded by arrows $A\stackrel \alpha\to B\stackrel \beta\to C$ and $A\stackrel \gamma\to D\stackrel \delta\to C$ is said to commute if $\delta\circ \gamma =\beta\circ\alpha$. In the diagrams with short exact sequences, you have two such meshes; the condition just described shall hold for both meshes. –  Hagen von Eitzen Nov 21 '12 at 11:56
thanks. this is helpful. i will work with this as a startpoint. hopefully i´ll get things done. –  grendizer Nov 21 '12 at 12:47
@HagenvonEitzen Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Sep 14 '13 at 10:40

A diagram with a mesh with vertices $A,B,C,D$ surrounded by arrows $A\stackrel \alpha\to B\stackrel \beta\to C$ and $A\stackrel \gamma\to D\stackrel \delta\to C$ is said to commute if $δ\circ γ=β\circ α$. In the diagrams with short exact sequences, you have two such meshes; the condition just described shall hold for both meshes.