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Can surjective functions map an element from the domain to two distinct elements in the codomain? I know that each element in the codomain must have at least one corresponding element in the domain. But for example,

$$f:\mathbb{R} \implies \mathbb{R^*} $$ $$f\left( \frac{a}{b}\right) = a + b $$

$$s.t.b \neq 0$$

this maps zero to every number, but it still is onto the codomain right?

By the way, $\mathbb{R^*}$ is the set of all real numbers not including zero.

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No. No surjective function can do that, because no function can do that. By definition, a function is required to map each element of the domain to a unique element of the codomain.

I recommend looking at the Wikipedia articles on functions, so-called "multivalued functions", and relations.

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  • $\begingroup$ what is it then? $\endgroup$ – user28823 Sep 8 '13 at 18:47
  • $\begingroup$ by the way, your "answer", does not qualify as an answer. $\endgroup$ – user28823 Sep 8 '13 at 18:47
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    $\begingroup$ You can call it a relation. Your $f$ shows why one must be careful when defining a function, because $f$ is not well-defined. This means that given an $x$ in the domain, $f(x)$ is not uniquely determined. $\endgroup$ – Stahl Sep 8 '13 at 18:48
  • $\begingroup$ @user28823 If you wanna read about it, functions are a special kind of (binary) relations. The entity you 'described' (it isn't properly defined) is a binary relation. $\endgroup$ – Git Gud Sep 8 '13 at 18:49
  • $\begingroup$ of course, $x_1 = x_2 \implies f(x_1) = f(x_2)$ for all x iff f(x) is a function, thank you $\endgroup$ – user28823 Sep 8 '13 at 18:50

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