Codomain

Image of a function(f) from X(left) to Y(right). The smaller oval inside Y is the image of f. Y is the Codomain of f.

"Definition" of codomain in informal mathematics is similar to this one:

Informal definition

In mathematics, the codomain, or target set, of a function, described symbolically as $f$ : $X$ → $Y$, is the set $Y$ into which all of the output of the function is constrained to fall.

All the output that the function can possibly produce from its given domain, $X$, is the image. The function's image will not necessarily fill the entire codomain $Y$, even though the output must all land inside of the codomain: there can be points in the codomain that are "not used."

The codomain (or target) is part of the definition of a function. The image (or range) is a consequence of the definition of a function: the image is a subset of the codomain and depends upon (i.e. is a consequence of) how the definition of the function prescribes the domain, codomain, and map or formula.

(The domain of $f$ is the set $X$.)

Examples

As an example, let the function $f$ be a function on the real numbers:

$f\colon \mathbb {R} \rightarrow \mathbb {R}$

defined by

$f\colon \,x\mapsto x^{2}$, or equivalently $f(x)\ =\ x^{2}$.

The codomain of $f$ is $\mathbb {R}$, but clearly f does not map to any negative number. Thus the image of f is the set

$\mathbb {R} _{0}^{+}$,i.e., the interval [0,∞) where:

$0\leq f(x)<\infty .$

We can define an alternative function $g$ thus:

$g\colon \mathbb {R} \rightarrow \mathbb {R} _{0}^{+}$

$g\colon \,x\mapsto x^{2}.$

While $f$ and $g$ map a given x to the same number, they are not, in the modern view, the same function because they have different codomains. To see why, suppose that we define a third function h:

$h\colon \,x\mapsto {\sqrt {x}}.$

We must define the domain of h to be $\mathbb {R} _{0}^{+}$:

$h\colon \mathbb {R} _{0}^{+}\rightarrow \mathbb {R}$.

Now let's define the compositions

$h\circ f$,

$h\circ g$.

As it turns out, $h\circ f$ doesn't make sense. Suppose (as we must, unless we explicitly state otherwise) that we do not know what the image of $f$ is; we only know that it can be $\mathbb {R}$. But then we are in trouble because the square root is not defined for negative numbers. Now we have a possible contradiction because function h, when composed on function f, might receive an argument which it "can't handle."

This unclarity should be avoided in formal work. Function composition therefore requires by definition that the codomain of the function on the right side of a composition (not its image, which is a consequence of the function and is said to be indeterminate at the level of the composition) must be the same as the domain of the function on the left side.

The codomain affects whether a function is a surjection. In our example, $g$ is a surjection while $f$ is not. The codomain does not affect whether a function is an injection.

A second example of the difference between codomain and image can be seen by considering the matrix of a linear transformation. By convention, the domain of a linear transformation associated with a matrix is Rⁿ and its codomain is R^m, where the matrix is $m\times n$ (has $m$ rows and $n$ columns) and the image is usually called the range. But the range (the set of numbers obtained when the matrix is right-multiplied by every column vector of length $n$) could be much smaller. For example, if the matrix contains only $0$s, then no matter how large it is, the range is just the vector 0. But the dimension of the resulting vector is $m$. This is important, because it is enough to change just one number in the matrix to make its range non-zero.

Formal aspects

The truth is that informal definition use nonstandard definition of function. Standard definition of function is the set of ordered pairs. So when having two functions, they are equal if and only if they have same elements, same pairs (Axiom of extensionality). So according to this definition functions:

$f\colon \mathbb {R} \rightarrow \mathbb {R}$

$f\colon \,x\mapsto x^{2}.$

$g\colon \mathbb {R} \rightarrow \mathbb {R} _{0}^{+}$

$g\colon \,x\mapsto x^{2}.$

are the same set, same function. As you might notice, formulas above are not true definitions, but each pair of these formulas is enough to create correct definition, so they are often used instead of it. First formula is equivalent to $function(f)\land dom(f)=\mathbb {R} \land rg(f)\subseteq \mathbb {R}$ and second is equivalent to $\forall {x\in dom{(f)}\;}{(f(x)=x^{2})}$. For function $g$ first formula is equivalent to $function(g)\land dom(g)=\mathbb {R} \land rg(f)\subseteq \mathbb {R} _{0}^{+}$, and second is similar to formula for function $f$. Correct definitions looks like that:

$f{\overset {\underset {\mathrm {def} }{}}{=}}\!\,\{x^{2}\colon \,x\in \mathbb {R} \}$

$g{\overset {\underset {\mathrm {def} }{}}{=}}\!\,\{x^{2}\colon \,x\in \mathbb {R} \}$

In standard definition of function one function can have many codomains (every set which has image of function as its subset is function's codomain). If we would like to use codomains in nonstandard way similar to informal definition we have to use for example definition of function as a pair of two sets, first is set of pairs from standard definition, and second is function's codomain. Also we could use some axiomatical approach like Category theory in which, among others, codomains are basic objects. But nonstandard approaches create some problems like our intuitions of function composition. In example above author claims that $h\circ f$ doesn't make sense. According to first order logic there is nothing like nonsense terms. (They are possible in other logics). Moreover composition has strict definition which goes well with our intuitions:

$h=f\circ g\iff \forall {x,y}\left[xhy\iff \exists {z}\left(xgz\land zfy\right)\right]$

Which means, that $y=(f\circ g)(x)$ if and only if there exist some element $z$ such that $z=g(x)$ and $y=f(z)$. As you might see, the "problems" claimed by the author of informal definition are nonexistent. If function $f$ from informal definition has some negative values there is no such z that is the element of domain of $h$.

Informal definition showed above is a great example of how informal mathematics allows some vague notions, and why formalization of mathematics is important issue.