Left Inverse and Surjectivness I just needed to clarify something.
I read the following proposition and something didn't make sense:
"The map $f$ is injective if and only if $f$ has a left inverse"
Now $f$ having a left inverse implies there is a function $g$ whose domain is the codomain of $f$.
Every element of the domain of $g$ must have a specified value (by definition), then surely $f$ is also surjective since the codomain of $f$ is its range.
My question is, does $f$ having a left inverse imply that $f$  is bijective?
 A: The left inverse does not prove surjectivity. Take $X = \{0,1,2\}, Y = \{0,1,2,3\}$ and $f(x) = x$. This has a left inverse $g(x) = x, g(4) = \text{anything in }\{0,1,2\}$ from $Y$ to $X$. The value 4 is not assumed, so it does not matter what value $g$ has there. What matters is that when we start with a value in $X$, we get in $f[X]$ and $g$ has to be defined there to go back to the (must be) unique $x$ again. Yes, $g$ must be defined on $Y$, but anything outside $f[X]$ is irrelevant. We only need $g(f(x)) = x$ for all $x$.
A: Your counterexample is the function $\exp:\Bbb R\to\Bbb R$, injective but not surjective. For its left inverse take $L:\Bbb R\to\Bbb R$ by: If $t>0$, $L(t)=\log t$ (natural logarithm of course), and if $t\le0$, then $L(t)=17$ (my favorite value).
You’ll notice that the domain of $L$ is the codomain of $\exp$ (not the range (“image”, in the language I grew up with)) and that the codomain of $L$ is the domain of $\exp$. That’s what’s necessary for $L$ to be a left inverse of $\exp$.
It’s true that every function becomes surjective when you shrink the codomain to the range of the function; that’s another issue.
