How exactly is a Field monomorphism defined? Multiple times in my question I'll use the term 'a mapping of $x$', what I mean with this is one of the element to which $x$ can be mapped to (since there could be more than one).
We have Two fields, $L$ and $K$, which operators are $+$ and $×$. A field monomorphism $α:K→L$ would be a homorphism, that is, a mapping that suffices the properties
$$α(x+y)=α(x)+α(y)$$
$$α(x×y)=α(x)×α(y)$$,
and is injective, that is, it never maps two different elements in $K$ to the same element in $L$. However, the possibility is open for an injective mapping to map the same element in $K$ to two, or more, different elements in $L$. This is what is confusing me.

Multiple times in my question I'll use the term 'a mapping of $x$', I refer with this phrase to one of the element to which $x$ is be mapped to by $α$ (since there could be more than one).

Imagine that $α$ maps $x$ to $r_1$ and $r_2$, and maps $y$ to $s_1$ and $s_2$. Then, by definition, 
$$α(x+y)=α(x)+α(y)$$,
but what does this exactly mean? Does it mean that for whichever mapping of $x$ and $y$ we pick, $α(x+y)$ must equal their sum, in the sense that $(x+y)$ would be mapped, at the very least to $(r_1+s_1), (r_1+s_2), (r_2+s_1)$ and $(r_2+s_2)$? 
Or does it mean that for each mapping of $x$, there has to be at least one mapping of $y$ such that $α(x+y)=α(x)+α(y)$, in the sense that $(x+y)$ would be mapped at the very least to $(r_1+s_1)$ and $(r_2+s_2)$? 
Or, my last interpretation, does it mean that there has to exists at least one mapping of $x$ and one mapping of $y$ such that $α(x+y)=α(x)+α(y)$, in the sense that $(x+y)$ would be mapped to at least $r_1+s_1$?

Which one is the correct interpretation? If nine of the interpretations I gave is correct, what would be the proper one?
I would really appreciate any help/thoughts!
 A: All field homomorphisms are monomorphisms.
All fields are rings, and we have that the kernel of any ring-homomorphsim is an ideal of the domain ring (verify this!).
If a ring ideal (of a ring $R$) $I$, contains a unit, say $u$, then $I = R$, because for any $r \in R$, we have $r = (ru^{-1})u \in I$.
It follows that if $R$ is a field, there are only two possible ideals: $R$ and $(0)$.
If $R=F$, a field, then any ring-homomorphism $\phi:F \to E$ image is thus:
$F/F \cong\{0\}$, or $\phi(F) \cong F/(0) \cong F$.
The first possibility cannot be a field-homomorphism, because in a field $0 \neq 1$, and the image is not a field, because it lacks a multiplicative identity distinct from $0$.
The second (and thus only) possibility is a monomorphism, because it has kernel $0$.
(In fact, the above argument can be extended to division-rings, which shows all division-ring homomorphisms are also monomorphisms. The condition that all non-zero elements are units, is thus very restrictive on what kind of ring-homomorphisms are possible, which essentially means the study of division-ring and field morphisms boils down to the concept of extensions, which serve as the main topic of investigation).

As others have commented above, it cannot happen for a function $f$ to have two distinct images $f(x)_1$ and $f(x)_2$-such a relation is called an ill-defined function, for then $f(x)$ is unclear as an element of the co-domain, although it may make sense as a set.
Since homomorphisms are required to be (well-defined) functions, the behavior you are concerned about cannot happen. This becomes particularly important when studying quotient objects, because a map (in rings, for example) $R \to R/I$:
$a \mapsto a+I$ has to be unambiguous, no matter which pre-image of the set $a+I$ we pick.
One of the consequences of this is that "quotient fields" are for all intents and purposes, nonexistent.
A: In mathematics, unless stated otherwise, the term "mapping" (or "map", or "function") refers to a single-valued operation.  In other words, if $f:X\to Y$ is a mapping, for each $x\in X$ there is exactly one value $f(x)\in Y$.  More formally, you can think of $f$ as being given by a set  $S$ of ordered pairs $(x,y)\in X\times Y$ such that for each $x\in X$ there is exactly one $y$ such that $(x,y)\in S$.  We then write $f(x)$ for the unique $y$ such that $(x,y)\in S$.
So, your $\alpha$ always maps any $x\in K$ to exactly one element of $L$, and there is no ambiguity of the sort you are asking about.
