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You can get a complex number when (for example) taking the root of a (negative) real number, but can you also do the same for quaternions? So my question is:
Is there an algebraic function $f$ that maps $\mathbb{C}$ (complex numbers) to $\mathbb{H}$ (quaternions)?

EDIT:
$f$ should map a number $v\in\mathbb{C}$ possibly $\in\mathbb{R}$ to a number $w\in\mathbb{H},\not\in\mathbb{C}$.
Also $f$ should not be trivial meaning the function shouldn't "contain" quaternions, example: $f(x)=x+a,a\in\mathbb{H}$ would be trivial. (Is there a better way to say what I mean?)

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    The quaternions $\mathbb{H}$ can be viewed as a strict non-commutative extension of $\mathbb{C}$. Indeed one may take $f(a+bi) = a+bi$, where $a$ is the real part and $b$ the imaginary part of any complex number. But you could send $i$ to $j$ or to $k$ in the quaternions with the same effect. – 2017-02-11
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    Could you explain what you mean with "send", please? – 2017-02-11
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    I mean function $f$ "sends" $a+bi$ in the complex numbers to $a+bi$ in the quaternions, in the sense that a function assigns a value in the range to each value in its domain. "sends" means "maps". So in the obvious way the complex numbers can be viewed as a subalgebra of the quaternions. – 2017-02-11
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    You could send $a+bi$ to $x(a+bi)x^{-1}$ for any quaternion $x$ (generalizing hardmath's example). – 2017-02-11
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    Thanks, but this seems trivial because $\mathbb{R}\in\mathbb{C}\in\mathbb{H}$ will edit the question to make this more specific. – 2017-02-11
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    It did seem trivial, which is why I bypassed posting an Answer. I look forward to your clarification of the Question. – 2017-02-11
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    Maybe the answer here is that there's no "natural" map on $\Bbb C$ that produces quaternions since $\Bbb C$ is algebraically closed – 2017-02-11
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    These kinds of embeddings of $\Bbb C$ in $\Bbb H$ are missing a feature in the OP's post: in the example given, we're "forced" to leave $\Bbb R$ by our desire to take square roots of negatives. But since $\Bbb C$ is algebraically closed, we'll need a different sort of trick to be forced to leave $\Bbb C$. I'm wondering if we can force our exit by asking for non-commutativity; perhaps seeking nonzero solutions to $iz = -zi$. – 2017-02-11
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    @pjs36 I indeed saw a different [question](http://math.stackexchange.com/questions/789451/do-there-exist-equations-that-cannot-be-solved-in-mathbbc-but-can-be-solve?rq=1) about an equation having it's solutions in $\mathbb{H}$ due to commutativity. – 2017-02-11

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As it turns out this isn't possible the way stated.