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Solución:
Creo que su confusión puede deberse a los diferentes usos de la terminología.
Para una función $f: X rightarrow Y$, la codominio es solo el conjunto $Y$. Por ejemplo, si $f: mathbbR rightarrow mathbbR$ es $x mapsto x^2$, entonces el codominio es $mathbbR$. Todos los que la usan están de acuerdo con esta terminología: es decir, nunca he visto a nadie usar el término “codominio” para significar otra cosa.
Desafortunadamente el término rango es ambiguo. A veces se usa exactamente como se usa el codominio arriba, por lo que algunos dicen que $mathbbR$ es el rango de la función de elevar al cuadrado definida arriba. Sin embargo, aquellos que usan el término codominio usualmente reservan el término “rango” para el subconjunto $y in Y $, es decir, el subconjunto de valores que en realidad están mapeados por algún elemento en el dominio. (Algunos otros usan el término imagen para esto en su lugar). Entonces, en el ejemplo anterior, la imagen de la función es $[0,infty)$. Whether the range is $mathbbR$ (i.e., the codomain) or $[0,infty)$ (i.e., the image) depends upon your convention, and both are rather prevalent.
In practice, this means that it would be safest never to use the term range, instead using codomain and image. (But most people don’t do that either…)
To add a bit to Pete’s and lhf’s answer: assuming we are talking about functions from one set to another, there are two common ways of viewing functions:
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As sets $f$ such that:
- Every element of $f$ is an ordered pair; and
- If $(x,y)$ and $(x,y’)$ are both elements of $f$, then $y=y’$.
When viewing functions this way, the domain is the set: $xmidtextthere exists ytext such that (x,y)in f$.
We write $f(x)=y$ to mean $(x,y)in f$. And two functions are equal if and only if they are equal as sets. In this case, when we write $fcolon Ato B$, we mean that $A$ is the domain, and that $B$ is any set such that $ymid textthere exists xtext such that (x,y)in f\subseteq B$. Note that the function $fcolon mathbbRtomathbbR$ and the function $gcolonmathbbRto1,-1$ given by $f(x)=g(x)=1$ if $xgeq 0$, $f(x)=g(x)=-1$ if $xlt 0$ are equal functions, because they are equal as sets.
One problem with this definition is that it makes no real sense to ask if a given function is “surjective”, because there is no understood “target set”. You have to ask about “surjective onto $B$“.
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As ordered triples $(A,B,f)$, where $A$ and $B$ are sets, and $fsubseteq Atimes B$ is a function in the sense above. $A$ is called the domain, $B$ is called the codomain, and two functions are equal if and only if they have the same domain, the same codomain, and the same value at each element of the domain. This is, I believe, the Bourbaki approach. In this case, the two functions $f$ and $g$ I mentioned above are different, because they have different codomains.
Here, the notion of surjective does make sense, and it provides good duality with the notion of one-to-one (though not complete duality; e.g., “a one-to-one function has a left inverse” only requires you to assume the domain is nonempty, but “an onto function has a right inverse” requires, and in fact is equivalent to, the Axiom of Choice).
This is what leads to the problems of the meaning of “codomain”, I think. The codomain of a function only makes sense (in the singular definite article way) in the second definition. The similar set that can be defined uniquely and precisely in the first definition is the image, which would be the dual of the domain: the set $ymidtextthere exists ztext such that (z,y)in f$. But, being the dual definition, there is a temptation of calling it the “codomain”, which clashes with the definition in the second instance.
A function $f: X to Y$ has $X$ as domain and $Y$ as co-domain. The image of $f$ is the subset of $Y$ given by $F(X)= f(x) : xin X subseteq Y$, that is, the set of values taken by $f$. When $F(X)=Y$, we say that $f$ is surjective. In general, $F(X)ne Y$. For example, $sin: mathbb R to mathbb R$ is not surjective but $sin: mathbb R to mathbb [-1,1]$ es.
Como se menciona en wikipedia, algunas personas usan rango por codominiootros para imagen. Creo que es mejor evitar rango en total.