Can't figure out the difference between them. I have read wiki article about codomains and images, but what is the difference? It seems confusing the examples part in codomain article. How can we claim this:

$f: \mathbb R \to \mathbb R$,

where $f(x) = x^2$? This function will never assume negative number, so why the codomain is R? After all, the authors might have gone even further and claim that codomain is the set of complex numbers!


1 Answer 1


You cannot read off the codomain from the formula $f(x)=x^2$.

Domain and codomain really are part of the data which comes with a function. This means that you cannot just say

"Let $f$ be the function $x\mapsto x^2$."

Instead, you always have to specify domain and codomain first, as in

"Let $f$ be the function from $\mathbb R$ to $\mathbb R$ mapping $x$ to $x^2$."

Or, as you mentioned, it could be

"Let $f$ be the function from $\mathbb R$ to $\mathbb C$ mapping $x$ to $x^2$."


"Let $f$ be the function from $\mathbb C$ to $\mathbb C$ mapping $x$ to $x^2$."

This will really be different functions.

Of course, if you want to define a function $f\colon X\to Y$ you have to make sure that $f(x)$ actually is an element in $Y$. Therefore,

"Let $f$ be the function from $\mathbb C$ to $\mathbb R$ mapping $x$ to $x^2$."

does not define a function.

The image of a function $f\colon X\to Y$ is, by the way, the subset of $Y$ consisting of all element $y\in Y$ for which there exists an element $x\in X$ with $f(x)=y$.

  • $\begingroup$ Thanks a lot, Rasmus! But isn't square of complex number a rational number? $\endgroup$ Commented Jan 8, 2012 at 15:13
  • 2
    $\begingroup$ @Zapadlo: No, it's not. The answer is also no, if you meant real instead of positive. The square of $1+i$, for instance, is not a real number. $\endgroup$
    – Rasmus
    Commented Jan 8, 2012 at 16:04
  • $\begingroup$ (replace real with rational in my previous comment) $\endgroup$
    – Rasmus
    Commented Jan 8, 2012 at 17:32
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    $\begingroup$ Are domain and images then same ? $\endgroup$ Commented Feb 1, 2017 at 18:06
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    $\begingroup$ @bodo For one thing, it might be extremely hard to determine the image of a given function. Let's consider a complicated curve in the plane. The codomain is just the plane because that's where the curve "lives". But writing down the list of points met by the curve is going to be messy and not something we want to do just to be able to define the curve. $\endgroup$
    – Rasmus
    Commented Jan 9, 2018 at 5:38

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