onto-functions

Mon Jun 29 2026

The whole question onto answers is simple: can every output be produced? Pick any $y$ you like in the codomain, and onto promises some input maps to it. Geometrically, for $f: \mathbb{R} \to \mathbb{R}$, the graph has to touch every height on the y-axis. No gaps, no unreachable $y$.

Proving onto: run the definition backwards

The definition reads "for every $y$, there exists an $x$", so the proof always has the same four moves. Start with an arbitrary $y \in B$. Solve $f(x) = y$ for $x$ in terms of $y$. Check that this $x$ actually lands in the domain $A$. Then plug it back and confirm $f(x)$ really simplifies to $y$.

For $f(x) = 2x - 3$ on $\mathbb{R} \to \mathbb{R}$, the candidate input is $x = (y+3)/2$, which is real for any real $y$, and substituting it back gives $y$. That is the recipe in full: find the preimage, check it lives in the domain, verify it maps where you want.

Not-onto, and what it does not mean

The cheapest not-onto function is $f(x) = 1$ on $\mathbb{R} \to \mathbb{R}$. Every input is used, but the output is just $\{1\}$, so the rest of $\mathbb{R}$ is never reached. The function $f(x) = x^2$ misses every negative number the same way. Both are perfectly valid functions, they just fail to cover their codomain.

Why onto is worth proving

Two reasons. First, guaranteed solvability. Onto is an existence theorem in disguise: if $f$ is onto, the equation $f(x) = y$ has a solution for every $y$. In linear algebra, a transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ being onto is exactly the statement that $Ax = b$ is solvable for every $b$.

Second, onto is half of what you need for an inverse. To build $f^{-1}: B \to A$, the inverse has to be defined on all of $B$, because a function is total on its domain. If $f$ is not onto, some $y \in B$ has no preimage, so $f^{-1}$ has nothing to send it to and fails to be a function.

Onto is only half of an inverse

Onto gives every $y$ at least one preimage. For an inverse you also need at most one, which is the one-to-one (injective) condition: no output is shared by two inputs, or $f^{-1}$ would have to send one $y$ to two places and stop being deterministic. Onto plus one-to-one is bijective, exactly one preimage per $y$, and that is when $f^{-1}$ exists. The one-to-one side, and the trap that catches most people proving it, is its own story in one-to-one-proof-converse-trap.

Equal sizes collapse onto and one-to-one

When the domain and codomain are finite and the same size, $|A| = |B| = n$, onto and one-to-one stop being separate properties. Prove one and the other comes free. The reason is the Pigeonhole Principle: if $f$ is not one-to-one, two inputs collide on one output, so the $n$ inputs cover at most $n-1$ outputs, leaving some element of $B$ unreached, which is not onto. Run it the other way and not-onto forces not one-to-one.

Infinite sets break this. Take $f: \mathbb{N} \to \mathbb{N}$ with $f(1) = 1$ and $f(n) = n - 1$ for $n > 1$. It is not one-to-one, since $1$ and $2$ both map to $1$, yet it is onto, since every $y$ has preimage $y + 1$. The bound that locked the two properties together is gone. This asymmetry is the very definition of an infinite set: a set is infinite exactly when it has a map to itself that is onto but not one-to-one (a Dedekind-infinite set).