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I have a question about this theorem and its proof:

Theorem

Let $f \in \mathcal{F}(A,B)$ be a function. $f$ is injective if and only if function $g \in \mathcal{F}(B,A)$ exists such that $g \circ f$ = ${I}_{A}$.

Proof

If $f$ is injective then for each $y \in Im(f)$ the set $f^{-1}(y)$ is an unitary set and the unique element of $f^{-1}(y)$ is denoted by $a_y$, it's true that $a_{f(x)} = x$. Let $x_0$ be a fixed element of $A$. We define: \begin{align*} g \colon B &\to A\\ y &\mapsto g(y) = \begin{cases} a_y & \text{if $y \in Im(f)$} \\ x_0 & \text{if $y \notin Im(f)$} \end{cases} \end{align*} Then, $g \circ f(x) = g(f(x)) = a_{f(x)} = x$ for each $x \in A$.

Reciprocally, if function $g \in \mathcal{F}(B,A)$ exists such that $g \circ f = I_A$, suppose that $x_1$, $x_2$ exists such that $f(x_1) = f(x_2)$, then $x_1 = (g \circ f)(x_1) = g(f(x_1)) = g(f(x_2)) = (g \circ f)(x_2) = x_2$.

Then $f$ is injective. QED.


Example

Example

What happens when $y \notin Im(f)$? In the example $white$ is not in the image set of $f$, my doubt is what really means that fixed $x_0$? what happens when you use $white$ as input in function $g$? $g(white) = {x_0}$?

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    $\begingroup$ $x_0$ is an arbitrary value of $A.$ In your example it can be $x_0=1,$ $x_0=2$ or $x_0=3.$ It will change the definition of the function $g$ but $g$ has not to be unique. $\endgroup$ – mfl Aug 8 '18 at 22:18
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I presume that your confusion is regarding the $(g \circ f$ = ${I}_{A}) \rightarrow (f \text{ is injective})$ part. In that case, the matter of $y$ such that $y$ is not in Im($f$) is immaterial. If you're trying to determine whether $f$ is injective, then all that matter is $a_0 \neq a_1 \rightarrow f(a_0) \neq f(a_1)$. The question of what happens when $y$ is not in Im($f$) doesn't matter, because it can't possibly affect whether $f(a_0) = f(a_1)$; both $f(a_0)$ and $f(a_1)$ are, by definition, in Im($f$). In your example, you can send $white$ to anything you want; it won't affect whether $f$ is injective. Whether $f$ is injective depends only on where the arrows in the left side of your diagram go. Where the arrows on the right side go is irrelevant.

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The theorem needs the hypothesis that $A$ isn't empty; it's false if $A=\varnothing\neq B$. But once we include that hypothesis, we can let $x_0$ be any element of $A$. There may be many possibilities for this $x_0$, but there is at least one, and that's all we need. Each such $x_0$ yields a function $g$ with the required property. So, from the fact that there is at least one $x_0\in A$ (because of the assumption that $A\neq\varnothing$), it follows that there is at least one $g$ as required.

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