A map from a nonempty set is one-to-one exactly when it splits on the left.Suggest a slogan
Let \(X\) be a nonempty set and \(f\colon X\to Y\) a map of sets. Then \(f\) is injective, that is, \(f(x) = f(x')\) implies \(x=x'\) for all \(x,x'\in X\), if and only if \(f\) has a left inverse: a map \(g\colon Y\to X\) with \(g(f(x))=x\) for every \(x\in X\).
If \(g\) is a left inverse and \(f(x)=f(x')\), then \(x=g(f(x))=g(f(x'))=x'\). Conversely, suppose \(f\) is injective and fix \(x_0\in X\). Define \(g(y)\) to be the unique \(x\in X\) with \(f(x)=y\) when \(y\in f(X)\), and \(g(y)\coloneqq x_0\) otherwise; then \(g(f(x))=x\) for every \(x\in X\).
The hypothesis \(X\neq\emptyset\) is needed: for a nonempty set \(Y\), the empty map \(\emptyset\to Y\) is injective but has no left inverse, since there is no map \(Y\to\emptyset\).
What condition on inverses characterises an injective function?