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Let A be a non-empty set and $f : A → A$ be a function.

Prove that f has a left inverse in $F_{A}$ if and only if f is injective (one-to-one).

$\leftarrow$ assume f is injective then $\forall x\in A \space \space \space \space \space \space \space \space \space f(x) \in A $ such that if $f(x)=f(y) $ then $ x=y$

something something $g(f(x)) = x \space \space \space \space \forall x\in A$

$\rightarrow$ assume f has a left inverse in $F_{A}$ then $\forall x\in A$

$g(f(x)) = x$ something says that x must be one to one?

Im really confused by this question First of all f must be a bijection if it is one to one from $ A \to A $ is it not?

Can someone help me out with this proof?

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    Are any functions $f : \{0\} \to \{1, 2\}$ one-to-one? Are they bijections too? Could you write down any left inverses for any of these one-to-one functions? – 2017-01-16
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    What is $F_A{}$? – 2017-01-16
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    it has a left inverse not a bijection not one to one. – 2017-01-16
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    $F_{A}$ is the set of functions that map A to A – 2017-01-16

2 Answers 2

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Suppose $f$ has a left inverse $g$.
Suppose $f(x)=f(y)$ for some $x,y\in A$.
Then $gf(x)=gf(y)$.
Since $gf=i_A$, we have $x=y$ and hence $f$ is injective.

Suppose $f$ is injective. Let $a_0$ be a fixed element in $A$.
Define $g:A\rightarrow A$ by the rules:
For $x\in A$
(i) If there is an element $y\in A$ such that $f(y)=x$, then $g(x)=y$.
(ii) If no such element $y$ exists in $A$, then $g(x)=a_0$.

For case (i), the element $y$ is unique as if $g(x)=y_1$ and $g(x)=y_2$, then $f(y_1)=x=f(y_2)$ which implies $y_1=y_2$.
For case (ii), $g(x)=a_0$ is also unique by the way we define it.
So $g$ is a function from $A$ to $A$.

For $a\in A$, $gf(a)=a$ as $f(a)=x\in A$ implies $g(x)=a$ by (i).
Hence $g$ is a left inverse of $f$.

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    ii) was the hard side. – 2017-01-16
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    $\forall y \in A g(f(y))=x $ so g compose f is the same as $ I (y) \forall y \in A$ – 2017-01-16
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    @Faust7 for all $y\in A$, $gf(y)=y$, so $gf=i_A$, the identity function from $A$ to $A$. – 2017-01-16
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    Yes that was what i was trying to say but couldn't :( Thank you very much your explanation was flawless. – 2017-01-16
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First of all, a one-to-one set function $f: A \to A$ certainly need not be a bijection; consider the multiplication by two map $\times 2: \mathbb{Z} \to \mathbb{Z}$. This is injective, but no odd numbers are in the image.

Now for your main question, assume $f$ is injective. Then define a left inverse $g$ by setting $g(x) = f^{-1}(x)$ wherever $f^{-1}(x)$ is defined. Because $f$ is injective, there is only one possible value for $f^{-1}(x)$, whenever it exists. Thus, $g$ is a left inverse of $f$ (because $g(f(x)) = x$ by definition). In the other direction, suppose that $g$ is a left inverse to $f$, i.e. for all $x \in A$, $g(f(x)) = x$. Now suppose that $f(x) = f(y)$. Applying $g$ to both sides, we see that $g(f(x)) = g(f(y))$, and since $g$ is a left inverse, this gives $x = y$. Thus $f$ is injective.