Proving a map is injective iff it has a left inverse

Question

Let $f:A\to B$

Prove the map $f$ is injective iff it has a left inverse.

Starting with $f$ is injective

Let $A$ not be $ \varnothing$

$f(A) = \{b \in B \ | \ b = f(a) \text{ for some } a \in A\}$

If there is a map $g: B \to A$ defined by $g(B) = \{a \in A \ | \ a = g(b)\text{ for some } b \in B\}$ where $b = f(a)$

The composite function $g \circ f(a): a \to f(a) \to a$ returns the identity on $A$ and is thus a left inverse.

$g$ does not exist in this definition if $g(b_i) \neq a_i$ for some $i$ denoting an element of $A,B$, which makes it the nullset

Or

$g(b_i) = a_i = a_{i+x}$ two or more elements of $A$ share an element of $B$ in the mapping of $f: A \to B$ It cannot be true because $a_i \neq a_{i+x}$ implies $f(a_i) \neq f(a_{i+x})$

Answer 1

Let $f$ be injective. We construct a left-inverse for $f:A\to B$.

Let $a\in A\neq\emptyset$ be an arbitrary element.

Now take $g: B\to A$, $b\mapsto \begin{cases} f^{-1}(b),\text{if $b\in\mathrm{Im}f$}\\ a,\text{else}\end{cases}$.

Where $f^{-1}(b)$ notes the single element of $f^{-1}(\{b\})$, as justified further below.

This function is well-defined, as every element $b\in B$ has an image in $A$, and the image is unique, so $b$ does not get mapped onto several different elements in $A$. Here the assumption that $f$ is injective comes in, as this implies that every element in the image of $f$ has exactly one preimage. So $f^{-1}(b)$ (which is the set (not to confuse with a function) of all elements in $A$ with $f(a)=b$).

Now we have $g(f(a))=a$ for every $a\in A$, as $f(a)$ clearly is an element in $\mathrm{Im}f$ (image of $f$).

So $g$ is a left-inverse of $f$.

For the converse:

Let $g$ be a leftinverse of $f$. We have to show that $f$ is injective.

So let $f(a)=f(a')$. We have to show that $a=a'$.

We have $g(f(a))=a$, as $g$ is a leftinverse. Also we get $g(f(a'))=a'$ for the same reason.

But $a=g(f(a))\stackrel{f(a)=f(a')}{=}g(f(a')=a'$. So $f$ is indeed injective.

Answer 2

Say $f$ is injective. Then for any $a, b$ in the domain of $f$, we have $f(a) = f(b) \implies a = b$. Thus, "the element that maps onto $x$" is uniquely defined for any $x$ in the codomain of $f$. Call this element $f^{-1}(x)$ - we see clearly that $f^{-1}(f(x)) = x$, so we have a well-defined left-inverse.

Say $f$ has a left-inverse $f^{-1}$. Let $f(a) = f(b)$. Applying $f^{-1}$ to both sides, we get $a = b$ (we know the left-inverse must be injective, or else $f$ would not be well-defined). Thus $f$ is injective.