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Can a function be surjective such that $\forall y \in Y$ there exists $x \in X$ such that $f(x) = y$ but such that not all $x$ values actually correspond to a $y$ value?

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    As long as every possible value of y is "corresponded to", then you have a surjective function. So in your case I would say your function is surjective. – 2017-01-09
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    @Kaynex now I have to contradicting answers? Your's and Cettt's answer? – 2017-01-09
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    The condition "not all $x$ correspond to some $y$" is in contradiction with the definition of function, which says that "all $x$ correspond to exactly one $y$". – 2017-01-09
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    @Crostul my function says that for all y there exist's and x such that f(x) = y but it doesn't say that for all x there exists a corresponding y? That's what I'm confused about? – 2017-01-09
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    What about the function $\ln(x)$? It is surjective on all real numbers, right? Or do we have to fix our domain to the positive reals? – 2017-01-09
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    @Kaynex $\ln:\mathbb{R}^+\to \mathbb{R}$ is how the natural logarithm is defined. The domain can't be all of the reals, because $ln$ does not map the negatives to any real number. It's not defined there. – 2017-01-09
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    Is there such a thing as a surjective function? – 2017-01-09

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A function $f:X\to Y$ is by definition a map that takes all values $x\in X$ and sends it to exactly one value $f(x)\in Y$. So in short no. However, if $f:X\to Y$ is surjective it is possible (in some cases) to define a surjective function $g:Z\to Y$ where $Z\subsetneq X$ such that $g(z)=f(z) ~\forall z\in Z$. This function is not defined on all of $X$ and so there does exist a $x\in X$ such that $g(x)$ is not defined.

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    @K.M. well you haven't defined what your case is exactly. If your function is defined $f:X\to Y$ , then no, because this by definition means that the function $f$ maps every element of $X$ to some element of $Y$. – 2017-01-09
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    Is this the case in my specified function since my function states that for all y there is a corresponding x that yields f(x) but it doesn't specify that there is an f(x) for all x's? – 2017-01-09
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    @K.M. I think you are confusing injectivity, surjectivity and the definition of a function here. By definition of a function, if $x$ is in the domain of the function $f$, then there must be an element $y$ in the co-domain such that $f(x)=y$. – 2017-01-09
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    Right, so that applies to a surjective function too I assume? – 2017-01-09
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    @K.M. it applies to all functions. A surjective function is a function for which every element $y$ in the co-domain there exists an element $x$ in the domain such that $f(x)=y$, which is the converse of my previous statement, which applies for all functions. A function is surjective if its co-domain equals its range. – 2017-01-09
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a function $f: X \to Y$ by definition assigns exactly one value $y=f(x) \in Y$ to all $x \in X$.

So, no it's not possible.