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A bijection is a function that is both injective and surjective. Each element of the codomain then has exactly one preimage:This unique-preimage property allows the inverse function $f^{-1}:B\to A$ to be defined on all of $B$.

Function composition is not always commutative. The reversed function composition may be undefined when the necessary domain of a function and codomain do not match. A counterexample using functions $\mathbb{R}\to \mathbb{R}$ avoids that issue: let $f(x)=x+1$ and $g(x)=2x$. ThenBoth function compositions exist, and their values differ at every real $x$.

There are $n(n-1)\cdots (n-k+1)$ ordered selections without replacement. Each unordered $k$-element subset occurs in exactly $k!$ orders, so the binomial coefficient isOne precise specification of its domain, codomain, and map isHere ${\mathbb{N}}_0=\{0,1,2,\dots \}$, and $0!=1$ includes the single empty selection. The actual values on $D$ are positive, so ${\mathbb{N}}_{+}$ is also a valid codomain. For fixed $n$, one can instead use the function $C_n:\{0,\dots ,n\}\to {\mathbb{N}}_{+}$ with $C_n(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$.

For $0\le k\le n$, the adjacent binomial coefficients obeySince $\left(\genfrac{}{}{0ex}{}{n}{k}\right)>0$ and $k+1>0$, the sign is determined by the numerator:The printed codomain ${\mathbb{N}}_{+}$ is inconsistent with the displayed map: $g(n)=-1$, and when $n$ is odd, $g((n-1)/2)=0$. With the usual extension $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$ for $k>n$, $g$ is a function ${\mathbb{N}}_0\to \mathbb{Z}$, and $g(k)=0$ for $k>n$; the positive range above remains unchanged. Without that extension, the natural domain is $\{0,\dots ,n\}$.^[ref]

The image of a function $f:A\to B$ isThe function is surjective, or onto, exactly when every $b\in B$ equals $f(a)$ for at least one $a\in A$, equivalently $f(A)=B$. It is injective, or one-to-one, exactly whenThus surjectivity concerns whether the entire codomain is reached; injectivity concerns whether distinct inputs have distinct outputs.

A function is bijective if it is both injective and surjective. Equivalently, each element of the codomain has exactly one preimage.

A function $f:X\to Y$ is surjective if every element of its codomain has a preimage: for every $y\in Y$, there is some $x\in X$ such that $f(x)=y$.

Not injective; surjective. The equal values $f(0)=f(2)=1$ disprove injectivity. For any $y>0$, take $x=1+y^{-1/2}>1$. This lies in the stated domain of a function, andConsequently every member of the codomain has a preimage.

For a function $f:X\to Y$ and a subset $U\subseteq Y$, its preimage isIt is a subset of the original domain of a function. The notation does not require an inverse function: the preimage of a singleton set can have no members, one member, or many members. In the finite example, $f^{-1}(\{0\})=\{a,b\}$, $f^{-1}(\{1\})=\{c\}$ and $f^{-1}(\{3\})=\varnothing$. The preimage of a singleton set is also its fiber of a function.

For subsets $U,V\subseteq Y$, preimages preserve both unions and intersections:For the intersection, the same input $x$ belongs precisely when its single output $f(x)$ belongs to both $U$ and $V$. There is no opportunity to choose different witnesses, which is what made the image counterexample fail. The identities also hold for arbitrary families of output subsets.

In the finite example, choose $U=\{0,1\}$ and $V=\{0,2\}$. ThenThe intersection of the two preimages is exactly the same set $\{a,b\}$. Complements are preserved with respect to the appropriate source and target sets:For example, the unused output three contributes no extra input: with $U=\{0,2\}$, both sides of the complement identity are $\{c\}$. See Lebl, Proposition 0.3.15: preimages of unions, intersections and complements.

This makes preimages useful for transporting conditions. The solution set of $g(x)=0$ is $g^{-1}(\{0\})$, and a zero set is a preimage of zero. Requiring two output conditions takes the intersection of their preimages. In topology, a continuous function pulls every open set back to an open set; a measurable function pulls the specified measurable sets in its codomain back to measurable sets in its domain of a function. These definitions use preimages because their set operations remain consistent even when the function is noninjective.

A real-valued function has codomain contained in the real numbers.

For a function $f:X\to Y$ and a subset $A\subseteq X$, its restriction $f{|}_A$ keeps the same values but allows only inputs in $A$. Restricting the domain of a function can remove collisions and permit an inverse function, as the nonnegative quadratic function example shows. This changes the allowed inputs; it is different from keeping the inputs and changing the codomain.

A scalar-valued function is a function whose codomain is a field, typically the real numbers or complex numbers. Its output is a single scalar, while its input can belong to an arbitrary set. For example, the square of the norm of a vector is real-valued. In a relativistic application, being scalar-valued alone does not establish Lorentz invariance: the transformation of the input and of the complete quantity must also be specified.

A self-map is a function whose domain of a function and codomain are the same set. It need not be a bijection: a constant function on a two-element set is a self-map that merges its inputs. Self-maps can always be composed with one another, and all self-maps of a specified set form its full transformation monoid.