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So I'm told that for a function $f:X\rightarrow Y$ to be surjective then

$$\forall y\in Y~,~\exists x\in X~,~f(x)=y,$$

so does this "$\exists$" imply more than one such $x$ can hit the same $y$? That is, am I right in thinking that

$$\forall y\in Y~,~\exists ! x\in X~,~f(x)=y$$

is an incorrect definition of surjectivity?

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  • $\begingroup$ You are absolutely correct. $\endgroup$ Jul 13, 2013 at 4:05

2 Answers 2

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That's right. Surjectivity of a function $f: X \to Y$ does not require that for all $y\in Y$, there exists a unique $x \in X$ such that $f(x) = y \in Y$.

We can even have $X = \mathbb N$, $\;Y = \{1\},\;$ with $f: X\to Y$, such that $f(x) = 1\; \forall x \in X$. And indeed, $f$ is thereby surjective.

You're second definition, FYI, defines a bijective function.

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Yes, you are correct. If the $x$ is unique, then $f$ is not only surjective, but it is also injective. So your second definition is actually the definition of a bijective function. Since there exist functions that are surjective but not injective (see amWhy's answer for an example), the second definition is incorrect (it is "too strong" of a definition).

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