How to prove a bijection is "well defined" and "surjective" properly? (with an example) Here is an example of my proof of a question:
Prove that $\mathbb{C}^*$ is isomorphic to the subgroup of $GL_2(\mathbb{R})$ consisting of matrices of the form.
$$
\begin{pmatrix}
a&b\\
-b&a\\
\end{pmatrix}.
$$
Here is my proof:
Let $z\in\mathbb{C}^*=\{z=a-bi; a^2+b^2\neq 0\}$
and
$g\in G=\{\begin{pmatrix}
a&b\\
-b&a\\
\end{pmatrix}
;a^2+b^2\neq 0\}
$.
Suppose there exists a mapping $\phi$, such that $\phi:\mathbb{C}^*\to G$ by $\phi: z\mapsto g$. This relation is well defined and it is onto because for every $z=a-bi\in\mathbb{C}^*$, we have the representation $\begin{pmatrix}
a&b\\
-b&a\\
\end{pmatrix}\in G$. It is also injective because when $\phi(z_1)=\phi(z_2)$, we have
$$ \phi(z_1)=
\begin{pmatrix}
a_1&b_1\\
-b_1&a_1\\
\end{pmatrix}
=
\phi(z_2)=
\begin{pmatrix}
a_2&b_2\\
-b_2&a_2\\
\end{pmatrix}.
$$
So $a_1-b_1i=a_2-b_2i$, in another word $a_1=a_2$ and $b_1=b_2$, therefore $z_1=z_2$.
Then to show that $\phi$ preserves the group operation, observe that
$$
\begin{align}
\begin{split}
\phi(z_1z_2)&=\phi((a_1a_2-b_1b_2)-(a_2b_1+a_1b_2)i)\\
\\
&=
\begin{pmatrix}
a_1a_2-b_1b_2&a_2b_1+a_1b_2\\
-(a_2b_1+a_1b_2)&a_1a_2-b_1b_2\\
\end{pmatrix}\\
\\
&=
\begin{pmatrix}
a_1&b_1\\
-b_1&a_1
\end{pmatrix}
\begin{pmatrix}
a_2&b_2\\
-b_2&a_2
\end{pmatrix}\\
\\
&=\phi(z_1)\phi(z_2).
\end{split}
\end{align}
$$
Hence $\phi$ is isomorphism and $\mathbb{C}^*\cong G$.
This is certainly not a hard question. And I would like you guys to comment on my proof and tell me if this is rigorous enough because, in my opinion, I found there is something unsatisfactory to me. For example, I want to explain why it is "well-defined" but I don't really know what "well-defined" means while both my textbook and my Prof always begin with "this function /map is well-defined" in their proof of bijection but rarely stating the reason. So I wonder if it is just a 'meaningless' convention for the proof of a bijection or there is something deeper underneath it?
My second unsatisfactions with this proof is about "onto/surjective". At this stage, I think I know what it means mathematically, which simply is that for every $y$ on the range we can find a unique $x$ on the domain that is mapped to it. However, when this statement is included in a proof, I immediately feel it lacks something so that as you can see from my example that I don't really prove it but instead restating or paraphrasing my previous definition of the map. This is also something that happened a lot during my reading of the textbook and my Prof's lecture. It sounds like as long as $\phi$ is "well-defined", then it must be "onto". I just want to ask is there anything I can do with the language here or actually, there is a brief proof for onto so that in any circumstances I can boost this sentence to a more convincing level?
These two questions have been around me from the beginning, and I know I haven't studied the very core part of this subject and probably never, but I just want to throw out these questions and see how you guys react to this.
Thanks.
 A: Let me give you a detailed commentary...

Let $z\in\mathbb{C}^*=\{z=a-bi; a^2+b^2\neq 0\}$
and
$g\in G=\left\{\begin{pmatrix}
a&b\\
-b&a\\
\end{pmatrix}
;a^2+b^2\neq 0\right\}
$.
Suppose there exists a mapping $\phi$, such that $\phi:\mathbb{C}^*\to G$ by $\phi: z\mapsto g$.

Okay; first, you are supposed to produce a bijection that is a group isomorphism. So saying "Suppose there exists a mapping $\phi$" doesn't do anything. You are saying "Suppose we have a function between the two sets". If you define it, then I don't have to suppose anything. If you don't define it, then you aren't doing anything except asking me to make an unwarranted assumption. So what you are trying to do is define a function, not "suppose that a function exists".
Second: the description of $\mathbb{C}^*$ says that it consists of all elements of the form $a-bi$ with $a^2+b^2\neq 0$ (note, it should specify that $a$ and $b$ are real numbers). Any such number. The description of $G$ says that it consists of all matrices of the form
$$\left(\begin{array}{rr}
a & b\\
-b & a\end{array}\right),$$
with $a^2+b^2\neq 0$ (again, the description is missing that $a$ and $b$ are expected to be real numbers).
But you can describe them with any letters; if I tell you that $G$ consists of all matrices of the form
$$\left(\begin{array}{rr}
\xi & \theta\\
-\theta & \xi\end{array}\right)$$
where $\theta,\xi$ are real numbers such that $\theta^2+\xi^2\neq 0$, then I've described the exact same set of matrices. I don't have to use $a$ and $b$: they are "dummy variables": their names don't matter in the description. You can change them at will to any two symbols that are distinct from each other and not being used elsewhere, and you will describe the exact same collection of objects. These letters are not fixed, so you should not assume them to be fixed.
You tell me: let $z\in\mathbb{C}^*$, and let $g\in G$. I can say: okay, I'm going to take $z=3-7i$; that's an element of $\mathbb{C}^*$. And I'm going to take $g$ to be the matrix
$$g = \left(\begin{array}{rr}
\pi & \sqrt{2}\\
-\sqrt{2} &\pi\end{array}\right).$$
That's a perfectly fine element of $G$.
Then you tell me that $\phi$ will send this $z$ to this $g$; well... okay... but what does it do with any other element of $\mathbb{C}^*$? And why am I sending this $z$ to this $g$? Note that you did not put any conditions on the $z$ and $g$ that I was supposed to pick, nor did you tell me that you were defining something for all $z\in\mathbb{C}^*$. The only thing you told me to do was to take an element of $\mathbb{C}^*$ and an element of $G$, and I did so.
Note that $a$ and $b$ in the descriptions of $\mathbb{C}^*$ and $G$ are not fixed numbers; they are free variables that can take any value as long as they satisfy the given conditions (both real, not both zero). Unless you tell me what they are, you aren't telling me what the function is.
So presumably you meant to do something like the following:


Define $\phi\colon\mathbb{C}^*\to G$ as follows: given $z\in\mathbb{C}^*$, write $z=a-bi$; then let $\phi(z)$ be the element
$$\phi(z)=g=\left(\begin{array}{rr}
a&b\\
-b&a
\end{array}\right).$$
Note that this $g$ indeed lies in $G$; and that since the expression of a complex number in the form $a-bi$ with $a$ and $b$ reals is unique, this completely determines the value of $\phi$ at $z$.


Moving on:

This relation is well defined and it is onto because for every $z=a-bi\in\mathbb{C}^*$, we have the representation $\begin{pmatrix}
a&b\\
-b&a\\
\end{pmatrix}\in G$.

It's not a "representation". What you mean is that the object you (tried but did not in fact) defined to be the value of $\phi$ at $z$ is indeed an object of $G$, which you should phrase as I did above or in some similar way. Note that here we are defining a function, so we need to check this. In your original phrasing, you instructed me to assume I had a function, so checking if it is a function is a waste. If you tell me "Assume you have a dime; now let's verify that you would then indeed have a dime", I would say "Well, if we are assuming I have a dime, why do we need to check that this would mean I have a dime? You just told me I should assume I do."
Moreover, you are doing the wrong thing to show it is onto/surjective. To show it is surjective, you must show that for every element $g\in G$ there is an element $z\in \mathbb{C}^*$ such that $\phi(z)=g$. But you start by picking an element of $\mathbb{C}^*$, and showing me that $\phi(z)$ lies in $G$. This only shows that you have correctly identified a domain and a codomain, not that $\phi$ is surjective. So your argument to show $\phi$ is surjective is incorrect.

It is also injective because when $\phi(z_1)=\phi(z_2)$, we have
$$ \phi(z_1)=
\begin{pmatrix}
a_1&b_1\\
-b_1&a_1\\
\end{pmatrix}
=
\phi(z_2)=
\begin{pmatrix}
a_2&b_2\\
-b_2&a_2\\
\end{pmatrix}.
$$
So $a_1-b_1i=a_2-b_2i$, in another word $a_1=a_2$ and $b_1=b_2$, therefore $z_1=z_2$.

This is fine, once you correctly define $\phi$.

Then to show that $\phi$ preserves the group operation, observe that
$$
\begin{align}
\begin{split}
\phi(z_1z_2)&=\phi((a_1a_2-b_1b_2)-(a_2b_1+a_1b_2)i)\\
\\
&=
\begin{pmatrix}
a_1a_2-b_1b_2&a_2b_1+a_1b_2\\
-(a_2b_1+a_1b_2)&a_1a_2-b_1b_2\\
\end{pmatrix}\\
\\
&=
\begin{pmatrix}
a_1&b_1\\
-b_1&a_1
\end{pmatrix}
\begin{pmatrix}
a_2&b_2\\
-b_2&a_2
\end{pmatrix}\\
\\
&=\phi(z_1)\phi(z_2).
\end{split}
\end{align}
$$

This is also perfectly fine.

Hence $\phi$ is isomorphism and $\mathbb{C}^*\cong G$.

If you had correctly shown that $\phi$ is onto, then this would indeed follow.
As for "well-defined"... see the comments the first three paragraphs this old answer of mine.
