# Why gives a commutative diagram a proof?

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?

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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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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
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?
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