Proposition: Suppose V and W are vector spaces. If a linear map $T:V \to W$ is bijective then it is invertible.
Proof(with comments): We need to construct a linear map $S: W \to V$ such that $ST=I _V$ and $TS=I_W$.
Define a map $S:W\to V$ such that $$S(T(v))=v \;for \;all\; v\in V.$$ We check to make sure that this $S$ "makes sense". Suppose $w\in W$. Thus there is a $v\in V$ such that $T(v)=w$ since $T$ is surjective. Thus $S(w)=S(T(v))=v$. Therefore $S$ is defined on all $W$. To check that $S$ is well defined suppose $v,v' \in V$ such that $T(v)=T(v')$. We want to prove that $S(T(v))=v=v'=S(T(v'))$. Since $T$ is injective then $S(T(v))=v=v'=S(T(v'))$. Thus $S:W\to V$ is well defined("makes sense").
Now to prove that $S:W\to V$ is a linear map. First we prove additivity. Suppose $w_1,w_2 \in W$ thus there exists a $v_1,v_2 \in V$ such that $T(v_1)=w_1$ and $T(v_2)=w_2$. Thus using definition of $S$, $S(T(v_1))=v_1$ and $S(T(v_2))=v_2$. Thus $S(w_1)+S(w_2)=S(T(v_1))+S(T(v_2))=v_1+v_2$. Since T is a linear map then $T(v_1)+T(v_2)=T(v_1+v_2)=w_1+w_2$. Thus $S(T(v_1+v_2))=S(w_1+w_2)=v_1+v_2$.Thus $S(w_1+w_2)=S(w_1)+S(w_2)$.
To prove homogenity, suppose $w\in W$ and $a\in F(field)$. Thus there is a $v\in V$ such that $T(v)=w$. Thus $aT(v)=T(av)=aw$ since $T$ is a linear map. Thus $S(aw)=S(T(av))=av=aS(T(v))=aS(w)$. Therefore $S$ satisfies both additivity and homogenity so $S$ is a linear map.
To prove $S$ is the inverse of $T$, we need to show $ST=I_V$ and $TS=I_W$. Suppose $v\in V$ thus $ST(v)=S(T(v))=v=I_V(v)$ so $ST=I_V$. Now suppose $w\in W$ thus there is a $v\in V$ such that $T(v)=w$. Thus $TS(w)=T(S(w))=T(S(T(v)))=T(v)=w=I_W(w)$ thus $TS=I_W$ so $S$ is the inverse of $T$. $\square$
My questions:
1) Is the way I defined $S$ correct ? I defined $S$ as a function then checked to make sure it "makes sense(well defined on domain $W$)" ? Also I defined $S$ in terms of $v\in V$ not $w \in W$ ? I saw some authors do this for example: Define a function $f:R\to R$ by $f(a+b)=a/2+b/2$. Then the authors check that this function $f$ they defined "makes sense".
2) Is the way I defined $S$ mathematically acceptable by other mathematicians ?
3) Is the proof correct ?
Best Answer
In short, the answers to your question are
What does well-definedness mean?
A function is defined in set-theory as a relation with a certain property which, back at the broad view, means one input value to the function gives one output value. Relations are an extremely useful concept to have in your toolkit, and essential terminology for any deeper work, but (as seen in the blog post comments above) many mathematicians neither think of nor write about functions at the set-theory level unless the context requires it.
At a higher level, when we write $f : X\rightarrow Y$, $f(\mathit{input}) = \mathit{output}$, then the function is well-defined if:
A lot of the time we don't need to check 1 and 3
We don't need to check 1 and 3 when we have a function defined by phrases like:
where the RHS depends only on $x$, or phrases like:
where the RHS depends only on $x,y,z$.
The parameters to $f$ range over their entire domains (all of $\mathbb R$ or $\mathbb R^3$), we don't need to check 1.
The form of $f$ means that for whatever particular values of $x$ (or $x,y,z$) we put into $f$, we only get one value coming out of the RHS expression, so we don't need to check 3.
Example: $\quad$ $f : \mathbb R \to \mathbb R$ with $f(x)=x^2-\pi$.
Here, there's no need to write anything about well-definedness (check 2 is too obvious).
Example: $\quad f : \mathbb R^2 \to \mathbb R~$ with $~f(x,x^2) = 3x~$ for all $~x\in\mathbb R$.
Example:
The definition is a little sloppy; the reader thinks "Where did $a$ and $b$ come from? $a+b$ has to be a real number, so I guess $a$ and $b$ are arbitrary real numbers". Let's assume so and apply our checks:
If I was writing this myself, I'd probably express it as follows.
I've left out the quantifiers for $a_1,a_2,b_1,b_2$, assuming the that reader knows how to play the game of well-definedness, and that the reader can immediately see any real can be written as the sum of two reals.
Your example: $\quad$ Define a map $S:W\to V$ such that $S(T(v))=v \text{ for all } v\in V$.