Let $G$ be a group and $+:G \times G \to G$ the addition on $G$. Let $$\phi:G \to G, g \mapsto g+g,$$ then, by definition, $\phi$ is a subset of $G \times G$ satisyfing $\forall g \in G \exists ! h \in G: (g,h) \in \phi$, where the object $h$ is denoted by $\phi(g)$. More precisely, it holds that $$\phi=\{(g,g+g)\ | \ g \in G\}.$$
Let $\mathbf{C}$ be the complex numbers $\mathbf{C}=\{a+ib \ | \ a,b \in \mathbf{R}\}$ and for example $$f:\mathbf{C} \to \mathbf{C}, \ a+ib \mapsto 2(a+ib).$$ My first question is, how is a map defined for "different names of objects" and is this problematic? More precisely, consider $e^{2\pi i} \in \mathbf{C}$. Then $e^{2\pi i}=1$ and thus $f(e^{2 \pi i})$ exists and is equal to $2$, despite it not being explicitly mentioned in the relation defined by $f$. This should not be problematic, however, when defining $f$ I am using something like a "complete set of representations" for $\mathbf{C}$ in order to describe all its elements, namely that all elements can be written as $a+ib$, right? This then makes it possible that I assigned a value to every possible expression of a complex number, despite it not being in the "canonical expression" $a+ib$.
When one now wants to check injectivity of a function $g:A \to B$, one has to show that for all $a_1,a_2 \in A$ satisfying $g(a_1)=g(a_2)$ it holds that $a_1=a_2$. One would proof this by letting $a_1,a_2$ be arbitrary elements of $A$ that satisfy this property. However, in the example above with $f:\mathbf{C} \to \mathbf{C}$ it would suffice to check this for all $a_1+ib_1,a_2+ib_2$ with $a_1,a_2,b_1,b_2 \in \mathbf{R},$ since any other expression that is contained in $\mathbf{C}$ is equal to such an expression and is thus mapped to the same object.
What confused me was the reason for the following: when regarding a function, say $h:\mathbf{Z} \to \mathbf{Z}$ and defining $x:=1$, why do I have a value assigned to $x$ and did not "forget to assign a value to this expression" $x$ or rather assign a value to all "new" symbols I can define (which would be infinitely many). The answer to that should then be the "complete set of representations" that then ignore the specific names of the objects as long as they are equal. Is my thought process correct or am I understanding something wrong?
Best Answer
When discussing a function $f : \mathbb{C} \to \mathbb{C}$, you asked: "how is a map defined for 'different names of objects' and is this problematic?" The answer is that the function is not defined for names of objects at all; it is defined for objects. The input to $f$ is a complex number, not a name of a complex number. "$e^{2\pi i}$" and "1" are two names for the same complex number. So if you write "$f(e^{2\pi i})$" or "$f(1)$" you are applying $f$ to the same number, so of course you get the same answer.
Perhaps what is bothering you is that you defined $f$ by writing $a+ib \mapsto 2(a+ib)$, so you seem to be assuming that the input to $f$ has been represented by an expression of the form "$a+ib$". You are implicitly using here the fact that every complex number has a unique representation in the form $a+ib$, so by specifying the output for all inputs of this form, you have unambiguously specified the value of $f$ for every complex number.
To see what can go wrong, suppose you tried to define a function $g : \mathbb{C} \to \mathbb{C}$ by saying $re^{i\theta} \mapsto re^{i\theta/2}$. Now your definition is ambiguous, because each complex number has multiple representations of the form $re^{i\theta}$. For example, $1 = 1e^{i0} = 1e^{i2\pi}$, so is $g(1)$ equal to $1e^{i0/2} = 1$ or $1e^{i2\pi/2} = -1$? The definition is ambiguous.
When proving that a function $g : \mathbb{C} \to \mathbb{C}$ is one-to-one, you start by letting $z_1, z_2 \in \mathbb{C}$ be arbitrary and assuming $g(z_1) = g(z_2)$, and then you have to prove that $z_1 = z_2$. Since every complex number can be expressed in the form $a+ib$, you can then write $z_1 = a_1 + ib_1$ and $z_2 = a_2+ib_2$ if you want. The uniqueness of the representation isn't important here, just the existence of such a representation. So you could also write $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$.