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The standard definition suggests that every element in the codomain should have a preimage. So, Can different elements in codomain or range have same domain? A worst question I think. Please reply...

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  • $\begingroup$ Koushik, yes, an "onto function" is implicitly a function, not some other kind of relation. $\endgroup$ – Thomas Andrews Feb 5 '16 at 15:19
  • $\begingroup$ Different elements in the co-domain cannot have the same element in the domain mapping to them. But you can have multiple elements of the domain map to the same element in the co-domain. But that has nothing to do with a function being onto. An onto function means every element in the co-domain has at least one element in the domain mapping to it. $\endgroup$ – Gregory Grant Feb 5 '16 at 15:26
  • $\begingroup$ An onto function is definitely a function. It's not like "tall Pygmy" or "trim sumo wrestler". $\endgroup$ – BrianO Feb 5 '16 at 15:36
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    $\begingroup$ There are a few occurrences of the word "function" where a modifying word actually makes it a non-function. "Multi-valued function," for example, and "partial function," and even "meromorphic function." @BrianO :) $\endgroup$ – Thomas Andrews Feb 5 '16 at 15:43
  • $\begingroup$ @ThomasAndrews with apologies, i delete my comment. Thanks $\endgroup$ – Shailesh Feb 5 '16 at 15:44
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A function $f\colon A\rightarrow B$ is a relation between the elements of $A$ and $B$ satisfying a key property: if $f(a)=b$ and $f(a)=b^{\prime}$, $b=b^{\prime}$.

That is, a function maps an input to only ONE element.

An onto function is a function $f\colon A\rightarrow B$ satisfying the following property: if $b$ is an element of $B$, there exists an $a$ in $A$ such that $f(a)=b$.

Intuitively, for each output, you can find an input that will produce that output.


Note that there could be more than one element mapping to each $b$ in $B$. For example, $f(a)=f(a^{\prime})=b$. If $C\subset B$, we can define the preimage of $C$ as $$ f^{-1}(C)\equiv\left\{ a\in A\colon f(a)\in C\right\} . $$ If $C$ is a singleton (i.e. $C=\{c\}$ for some $c\in B$), we often drop the set braces $\{\cdot\}$: $$ f^{-1}(c)\equiv f^{-1}(\{c\}). $$ We can now restate the definition of onto in terms of preimages of singletons:

An onto function is a function $f\colon A\rightarrow B$ satisfying the following property: for each $b$ in $B$, $f^{-1}(b)$ is nonempty.

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