I'm told that $g$ is a left inverse of $f$ if $g\circ f=1_X$. I'm also told that if $f$ has a left inverse, then $f$ must be injective. I'm now asked to prove the converse, namely that if $f:X\rightarrow Y$ is injective, then there exists a function $g:Y\rightarrow X$ such that $g\circ f =1_X$. Where might I start here?
[Math] Left Inverse: An Analysis on Injectivity
calculusfunction-and-relation-compositionfunctionsreal-analysis
Best Answer
One way to think about injective functions is that they preserve information in the domain. Suppose $f:X\to Y$ is injective, then we can tell whether $x=y\in X$ just by looking at whether $f(x)=f(y)\in Y$, so we see the information about points in the domain are preserved by this mapping.
Then it follows easily that such functions have left-inverses, because since all information in the domain is preserved in the image, then we can get the domain from the image. This is what a left-inverse does.
Following this philosophy, we might just define this $g: Y\to X$ on two pieces. On $f(X)$, we define $g(y)=x$ where $f(x)=y$. On $Y\backslash f(X)$, we just pick an arbitrary point $x^*\in X$ and let $g$ maps constantly to this $x^*$.