TITLE: Why gives a commutative diagram a proof?
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: @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?