Functions and invertibility

A function $f : A \to B$ is a relation from $A$ to $B$ in which every element of $A$ is linked to exactly one element of $B$. $A$ is the domain, $B$ is the codomain, and the range is the actual set of outputs achieved, which may be a proper subset of $B$. Four labels classify how a function behaves between its domain and codomain:

• One-one (injective): distinct inputs always give distinct outputs — $f(a_1) = f(a_2)$ forces $a_1 = a_2$.
• Many-one: the opposite — at least two distinct inputs share one output.
• Onto (surjective): every element of $B$ is hit by some input, so the range fills the whole codomain.
• Into: the opposite — some element of $B$ is never reached.

A function that is one-one and onto at once is a bijection, and bijection is precisely the condition for invertibility. When $f : A \to B$ is a bijection, define its inverse by reversing every arrow:

where: $a \in A$, $b \in B$, and $f^{-1} : B \to A$ is itself a genuine function only because bijectivity guarantees each $b$ has exactly one such $a$. The inverse satisfies $f^{-1}(f(a)) = a$ for every $a \in A$ and $f(f^{-1}(b)) = b$ for every $b \in B$ — applying $f$ then $f^{-1}$, or $f^{-1}$ then $f$, always returns you to the start.

A fast algebraic test for one-one: show that $f(x_1) = f(x_2)$ forces $x_1 = x_2$, or equivalently confirm $f$ is strictly monotonic (always increasing, or always decreasing) across its whole domain — a strictly monotonic function can never repeat an output.

Let $f : \mathbb{R} \to \mathbb{R}$ be $f(x) = x^2$. Decide whether $f$ is invertible, and if not, find a restriction that makes it invertible.

Step 1. Test one-one on $\mathbb{R}$. Take $x_1 = 2$ and $x_2 = -2$. Then $f(2) = 4 = f(-2)$ even though $2 \neq -2$, so $f$ is not one-one — it is many-one.

Step 2. Test onto on $\mathbb{R}$. Take $y = -1$ in the codomain. No real $x$ satisfies $x^2 = -1$, so $f$ is not onto — it is into. So $f : \mathbb{R} \to \mathbb{R}$ is neither one-one nor onto, hence not invertible.

Step 3. Restrict to $g : [0,\infty) \to [0,\infty)$, $g(x) = x^2$. For $x_1, x_2 \geq 0$, if $x_1^2 = x_2^2$ then $(x_1-x_2)(x_1+x_2)=0$; since both are non-negative, either both are zero or $x_1+x_2 > 0$ forces $x_1 = x_2$ — either way $x_1 = x_2$, so $g$ is one-one. For any $y \geq 0$, $x = \sqrt{y} \geq 0$ satisfies $g(x) = y$, so $g$ is onto.

Step 4. Conclude. $g$ is a bijection, hence invertible, and solving $y = x^2$ for $x \geq 0$ gives $x = \sqrt{y}$, so $g^{-1}(x) = \sqrt{x}$ for $x \in [0,\infty)$. The same formula $x^2$ that failed to be invertible on all of $\mathbb{R}$ becomes invertible once domain and codomain are restricted.

• Writing down an algebraic "inverse formula" without first checking that $f$ is actually a bijection on the stated domain and codomain — a formula that looks valid can still be meaningless outside a restricted range.
• Confusing "into" (range is a proper subset of the codomain) with "not a function" — an into function is still a perfectly valid function, just not onto.
• Classifying a function as one-one or onto without stating domain and codomain — as $f(x) = x^2$ shows, the same rule can flip its verdict depending on what sets are chosen.
• Applying the horizontal-line test backwards: one-one fails when a horizontal line crosses the graph more than once, onto fails when a horizontal line at some codomain height never meets the graph at all.

Ask the tutor: Not sure whether a given algebraic function is one-one, onto, or invertible on its stated domain — ask the AI tutor to walk through the check with you step by step.