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I am thinking about a proof of the following:

Suppose a map $f: A \to B$ has a retraction. Then for any set $T$ and for any pair of maps $x_1 : T \to A$, $x_2 : T \to A$ from any set $T$ to $A$ $$ \textrm{if } f \circ x_1 = f \circ x_2 \textrm{ then } x_1 = x_2. $$ The proof uses the diagram from the picture and I am wondering in what way the diagram shows anything? I understand the algebraic manipulations, but where does it follows from the diagram alone?

enter image description here

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I agree that the algebraic manipulation is much easier to understand (and to come up with). –  Martin Brandenburg Feb 5 '13 at 19:04
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up vote 13 down vote accepted

enter image description here ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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Yeah, colored diagram chases get 1+. –  Martin Brandenburg Feb 5 '13 at 19:04
    
i am still note sure... can I say that in general, if two paths meet at an object (in our case A) and if everything that lies behind that object (i.e. every path that starts from there) commutes, then it must be the case that the two paths which meet at that object are equal? –  Stefan Feb 5 '13 at 19:49
    
by equal paths I mean that the arrows which result in composition along the paths are equal. –  Stefan Feb 5 '13 at 19:50
    
@Stefan, I don't understand your question- what exactly do you refer to? –  user58512 Feb 5 '13 at 19:55
    
@user58512: see my post. –  Stefan Feb 5 '13 at 20:12
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@user58512

I am not sure $x_1$ equals $x_2$ because everything behind $A$ commutes. In analogy, because everything behind $C$ commutes, follows that $i\circ h = f \circ g$ in the following diagram?

enter image description here

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you said "everything commutes" which to me means i = f, h =g and C->D->F = C->E->F. This obviously implies hi = gf (you have written your composition backwards) for the same reason 1+1=2 and 4=2^2 implies (1+1)*4 = 2*2^2. Does that help explain or am I misunderstanding your question? –  user58512 Feb 5 '13 at 20:17
    
The reason the step from r f x_1 to r f x_2 is justified is only because we have f there and we have as an assumption f x_1 = f x_2. Without getting f involved we can't deduce x_1 = x_2: That's why we need a retract. –  user58512 Feb 5 '13 at 20:18
    
ok, but where in your diagram chasing is the fact used that $f\circ x_1 = f\circ x_2$? Your paths doesn't depend on this path (which is the bottommost arrow)... –  Stefan Feb 5 '13 at 21:44
    
the step diagram 3 to diagram 4 is justified by f x1 = f x2. –  user58512 Feb 5 '13 at 21:46
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ahhh, now I see it! –  Stefan Feb 5 '13 at 22:23
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