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From the book principle of mathematical analysis define 2.1:

Consider two sets $A$ and $B$, whose elements may be any objects whatsoever, and suppose that with each element $x$ of $A$ there is associated, in some manner, an element of $B$, which we denote by $f(x)$. $f$ is said to be a function from A to B(or a mapping of A into B )

My understanding: when we talk about a function, formally, we have to specify the domain $A$ and the codomain $B$.

Example:

  1. the function $f(x)=2x$, with the domain A=[1,2] and codomain B=[2,4]
  2. the function $f(x)=2x$, with the domain A=[1,2] and codomain B=[0,100].
  3. Strictly, formally, that is two different functions, right?
  4. Actually the former one is a surjective function, and the latter is not, right?
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  • $\begingroup$ See this discussion Basically, there are two ways of thinking about a function, leading to slightly different issues. $\endgroup$ – Arturo Magidin May 23 '19 at 17:47
  • $\begingroup$ In your example, isn't $A$ the domain and $B$ the codomain? As for your examples, your assertions are correct. $\endgroup$ – John Douma May 23 '19 at 19:03
  • $\begingroup$ See also math.stackexchange.com/q/60365. $\endgroup$ – Paul Frost May 23 '19 at 23:18
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To a set theorist the examples are the same function. In set theory a function IS its graph (as set theorists want everything to be a set). With this definition, a function can be said to be injective/surjective/bijective from its domain to a specified co-domain.

Your example is surjective to $[2,4]$ but not to any other set.

Anther term is the image of a function $f$, which is $\{y: \exists x\,(\,(x,y)\in f\}$. Note that "$\in f \,$" means "in the graph of $f $ ".

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  • $\begingroup$ Concerning the set-theoretic concept of a function you are right - unfortunately ;-) I would prefer to define a function as triple $(X,Y,f)$, where $X,Y$ are sets and $f \subset X \times Y$ is a subset with suitable properties. $\endgroup$ – Paul Frost May 23 '19 at 23:24
  • $\begingroup$ so, if we define a function as a triple $(X, Y, f)$, then $X$ is the domain, $Y$ is the image, and the $f$ is the rule? In this case, we don't care about the codomain that much, and think it as an additional thing of the function. $\endgroup$ – dawen May 24 '19 at 16:04
  • $\begingroup$ @dawen: No, generally in that formulation, $Y$ is the codomain, a set that contains (but need not be equal to) the image. $\endgroup$ – Arturo Magidin May 24 '19 at 17:31
  • $\begingroup$ Don't use math mode for italics; math italic is different from regular italic, and hard to read. To get italics, use *: e.g., writing *this* will produce *italics*, and two asterisks yields **boldface** gives: writing this will produce italics, and two asterisks yields boldface. $\endgroup$ – Arturo Magidin May 24 '19 at 17:40
  • $\begingroup$ OK, so, actually, the domain\codomain\image(which is implicitly in the $f$)\ everything is inside that triple formulation $(X, Y, f)$. And if we just consider the $f$, my example is actually the same function, but if we use the triple formulation, it is not the same function. $\endgroup$ – dawen May 24 '19 at 18:35

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