I've already done a lot of searching (in particular: https://www.proofwiki.org/wiki/Injection_iff_Left_Inverse) to try to prove this statement:

$f: A \to B$ is injective if and only if it has a left inverse. [I'm going to also assume $A$ and $B$ are nonempty.]

My attempt so far:

$\Leftarrow$ Suppose $f$ has a left inverse and $f(a_1) = f(a_2)$ for $a_1, a_2 \in A$. Then there is a function $g: B \to A$ such that $g\left(f(a_1)\right) = a_1$ and $g\left(f(a_2)\right) = a_2$. Since $g$ is well-defined, it follows that $g\left(f(a_1)\right) = g\left(f(a_2)\right)$, and hence $a_1 = a_2$.

$\Rightarrow$Now suppose $f$ is injective. Then $f(a_1) = f(a_2) \implies a_1 = a_2$. Furthermore, since $A$ is nonempty, there is an element $x \in A$. Define $g: B \to A$ by \begin{equation*} g(b) = \begin{cases} f^{-1}(b), & b \in \text{Im}(f) \\ x, & b \in A \setminus \text{Im}(f)\text{.} \end{cases} \end{equation*} This mapping is well-defined, since if $f$ is injective, $f^{-1}(b)$ is unique.

My main question: does this imply that if $f: A \to B$ is injective that any mapping $g: \text{Im}(f) \to A$ is bijective? Otherwise, I don't understand why we can even use $f^{-1}$ on an element. [Please read the link above for more details - in proof 1.]

  • $\begingroup$ What is 'left inverse'? $\endgroup$ – brick Dec 20 '14 at 20:39
  • $\begingroup$ @brick: $f$ has a left inverse if there is a function $g: B \to A$ such that $g \circ f: A \to A$ is the identity map on $A$, i.e., $(g \circ f)(a) = a$ for all $a \in A$. $\endgroup$ – Clarinetist Dec 20 '14 at 20:39
  • $\begingroup$ If it bothers you to say “$f^{-1}(b)$ is unique”, you can say instead “There is a unique $b'$ such that $f(b') = b$.” Or you can add a sentence before the definition of $g$ that says “Because $f$ is injective, for each $b$ there is a unique $b'$ such that $f(b') = b$; we will denote this $b'$ as $f^{-1}(b)$.” $\endgroup$ – MJD Dec 20 '14 at 20:42

does this imply that if $f:A→B$ is injective that any mapping $g:{\rm Im}(f)→A$ is bijective?

No. It only means that $f: A \to f(A) = {\rm Im}(f)$ is bijective. So you can consider the inverse, but with its domain restricted to the image of the initial function. So you can have more than one left inverse. The big theorem is that if exists both the left and right inverses, then they're equal. Any injective function is a bijection between its domain and its image. It is not true that any mapping $g: {\rm Im}(f) \to A$ is bijective.

I don't understand why we can even use $f^{−1}$ on an element.

In general you can't. Note that we only use $f^{-1}$ where it is well-defined, that is: in the image of $f$. It is well defined because $f$ is injective.

How would you go about showing that $f:A→B$ is injective $\implies f:A→{\rm Im}(f)$ is bijective?

Let's do all the details. Formally, two functions are equal if and only if all the domains, codomains, and rules of association are equals. Let $f:A \to B$ be an injective function. Consider $\bar{f}:A \to {\rm Im}(f)$ be defined by $\bar{f}(x) := f(x)$, for all $x \in A$. I'm going to prove that $\bar{f}$ is bijective.

For injectivity, take $x,y \in A$ such that $\bar{f}(x) = \bar{f}(y)$. So: $$\bar{f}(x) = \bar{f}(y) \implies f(x) = f(y) \implies x = y,$$ this last step being because $f$ is assumed injective.

For surjectivity, let $y \in {\rm Im}(f)$. By definition of image, exists $x \in A$ such that $f(x) = y$. But that means that $\bar{f}(x) = y$, so $\bar{f}$ is surjective.

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  • $\begingroup$ How would you go about showing that $f:A \to B$ is injective $\implies$ $f: A \to \text{Im}(f)$ is bijective? $\endgroup$ – Clarinetist Dec 20 '14 at 20:58
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    $\begingroup$ I'll edit this into my answer, give me two minutes (: $\endgroup$ – Ivo Terek Dec 20 '14 at 21:01
  • $\begingroup$ Done, please check it. $\endgroup$ – Ivo Terek Dec 20 '14 at 21:07
  • $\begingroup$ Thank you very much. So after showing that $\bar{f}$ is bijective, we could write [as a modification of my above question]: $$ g(b) = \begin{cases} \bar{f}^{-1}(b), & b \in \text{Im}(f) \\ x, & b \in A \setminus \text{Im}(f) \end{cases}$$ and this is clearly well-defined. To finish the proof off, just find $(g \circ f)(x)$ for all $x \in A$. Is my approach correct? $\endgroup$ – Clarinetist Dec 20 '14 at 21:11
  • $\begingroup$ Yes, it would be correct. But writing $\bar{f}$ is just an extreme detail of my part, most people wouldn't do this. Your proof wouldn't be criticized if you wrote $f$ straightforwardly. I chose to open up the details to help your understanding. $\endgroup$ – Ivo Terek Dec 20 '14 at 21:14

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