Injective, surjective, bijective কী?

Example 1.25 (Computing an inverse in practice)

Let $f \colon \R \to \intoo0{+\infty}$, $f(x) = \eu^{2x+1}$. To invert, solve $y = f(x)$ for a given $y > 0$:

each step being reversible on the announced domains. The computation delivers everything at once: for each $y$ in the codomain there is exactly one solution $x$, so $f$ is bijective, and

A quick check of both compositions ($f^{-1}(f(x)) = \frac{(2x+1) - 1}2 = x$ and $f(f^{-1}(y)) = \eu^{\ln y} = y$) confirms Theorem 1.24’s criterion. The insight: “solve for $x$ and watch the equivalences” is simultaneously the existence proof, the uniqueness proof, and the formula — but it only works if the codomain was announced correctly ($f$ is not surjective onto $\R$).

Example 1.27 (Point (2) is sharp)

In Proposition 1.26 (2), one cannot upgrade the conclusions: $g \circ f$ bijective does not force $f$ surjective or $g$ injective. Take $E = G = \{1\}$, $F = \{1, 2\}$, with $f(1) = 1$ and $g(1) = g(2) = 1$: then $g \circ f = \mathrm{id}_E$ is bijective, yet $f$ misses the element $2$ and $g$ collapses both elements. The moral is a precise bookkeeping rule: composition information flows to the inner map for injectivity and to the outer map for surjectivity, never the other way. (Exercise 1.9 builds the same phenomenon with infinite sets, where it is the engine behind one-sided inverses.)

Example 1.28

$f \colon \R \to \R$, $x \mapsto x^2$ is neither injective ($f(-1) = f(1)$) nor surjective ($-1$ has no preimage). Restricting the domain and codomain, $f \colon \R_+ \to \R_+$, $x \mapsto x^2$ is bijective, with inverse $y \mapsto \sqrt y$. A map’s injectivity or surjectivity depends on the announced domain and codomain, not only on the formula.