Maps (Functions)¶
Definition¶
A map (or function) \(f: A \to B\) assigns to each element \(a \in A\) exactly one element \(f(a) \in B\).
- \(A\) is the domain.
- \(B\) is the codomain.
- \(f(A) = \{f(a) : a \in A\} \subseteq B\) is the image.
Example. \(f: \mathbb{Z} \to \mathbb{Z}\) with \(f(x) = x^2\). Here \(f(3) = 9\) and \(f(-3) = 9\).
Injectivity¶
A map \(f\) is injective (one-to-one) if distinct inputs produce distinct outputs:
Example. \(f(x) = 2x\) is injective: from \(2a = 2b\) it follows that \(a = b\).
Counterexample. \(f(x) = x^2\) on \(\mathbb{Z}\) is not injective: \(f(3) = f(-3) = 9\), but \(3 \neq -3\).
Surjectivity¶
A map \(f: A \to B\) is surjective (onto) if every element in \(B\) has at least one preimage:
Equivalently: \(f(A) = B\).
Example. \(f: \mathbb{Z} \to \mathbb{Z}\) with \(f(x) = x + 1\) is surjective: for every \(b \in \mathbb{Z}\), \(a = b - 1\) is a preimage.
Counterexample. \(f: \mathbb{Z} \to \mathbb{Z}\) with \(f(x) = x^2\) is not surjective: \(-1\) has no preimage, since \(x^2 \geq 0\) for all \(x\).
Bijectivity¶
A map is bijective if it is both injective and surjective. Every element in \(B\) then has exactly one preimage.
Bijective maps possess an inverse map \(f^{-1}: B \to A\) with \(f^{-1}(f(a)) = a\) and \(f(f^{-1}(b)) = b\).
Example. \(f: \mathbb{R} \to \mathbb{R}\) with \(f(x) = 2x + 1\) is bijective. The inverse is \(f^{-1}(y) = \frac{y - 1}{2}\).
Composition¶
The composition of two maps \(f: A \to B\) and \(g: B \to C\) is the map \(g \circ f: A \to C\) with:
Example. \(f(x) = x + 1\) and \(g(x) = x^2\). Then \((g \circ f)(3) = g(f(3)) = g(4) = 16\).
Order matters: \((f \circ g)(3) = f(g(3)) = f(9) = 10 \neq 16\).
Summary¶
| Property | Meaning |
|---|---|
| Injective | Distinct inputs → distinct outputs |
| Surjective | Every \(b \in B\) has a preimage |
| Bijective | Injective and surjective; inverse exists |
| \(g \circ f\) | Composition: first \(f\), then \(g\) |
References¶
- Hammack, Richard: Book of Proof. 3rd edition, 2018. Chapter 12.