Problems with proof of a theorem about invertible linear transformations 
Theorem 10.2. A linear map $T : U \to V$ is invertible if and only if
  $T$ satises the equivalent conditions of Corollary 9.7. In particular,
  if $T$ is invertible, then $\dim(U) = \dim(V)$, so only square
  matrices can be invertible.
Proof: If any function $T$ has a left and right inverse, then it must
  be a bijection. Hence $\ker(T)=\{0\}$ and $\operatorname{im}(T) = V$ ,
  so $\operatorname{nullity}(T) = 0$ and $\operatorname{rank}(T) =
> \dim(V) = m$. But by Theorem 9.6, we have $n = dim(U) =
> \operatorname{rank}(T) + \operatorname{nullity}(T) = m + 0 = m$ and we
  see from the definition that $T$ is non-singular.
Conversely, if $n = m$ and $T$ is non-singular, then by Corollary 9.7
  $T$ is a bijection, and so it has an inverse $T^{-1} : V \to U$ as a
  function. However, we still have to show that $T^{-1}$ is a linear
  map. Let $v_{1}, v_{2} \in V$. Then there exist $u_{1}, u_{2} \in U$
  with $T(u_{1}) = v_{1}, T(u_{2}) = v_{2}$. So $T(u_{1}+u_{2}) =
> v_{1}+v_{2}$ and hence $T^{-1}(v_{1}+v_{2}) = u_{1}+u_{2}$.
  If $\alpha \in K$, then $T^{-1}(\alpha v_{1}) = T^{-1}(T(\alpha
> u_{1})) = \alpha u_{1} = \alpha T^{-1}(v_{1})$, so $T^{-1}$ is linear,
  which completes the proof.

As a reference I provide Theorem 9.6 and Corollary 9.7:

Theorem 9.6 (The rank-nullity theorem). Let $U, V$ be vector spaces
  over $K$ with $U$ finite-dimensional, and let $T : U \to V$ be a
  linear map. Then $\operatorname{rank}(T) +\operatorname{nullity}(T) =
> \dim(U)$.
Corollary 9.7. Let $T : U \to V$ be a linear map, and suppose that
  $\dim(U) = \dim(V) = n$. Then the following properties of $T$ are
  equivalent:
(i) $T$ is surjective;
(ii) $\operatorname{rank}(T) = n$;
(iii) $\operatorname{nullity}(T) = 0$;
(iv) $T$ is injective;
(v) $T$ is bijective;

Now I've got the following questions:
I'm not sure why the authors speak of left and right inverses in the proof of the theorem, but those have actually not been mentioned before. However. from my understanding we could replace "If any function $T$ has a left and right inverse" by "If a linear map $T$ is invertible". Also from what I understand the first part of the proof should establish the forward direction, i.e. show that if $T$ is invertible, it satisfies all of the properties (i)-(v) given in Corollary 9.7. To show this we need can show that $n=m$ and that $T$ satisfies one of those properties. The others then follow by Corollary 9.7. But what really confuses me in the proof is the last part of the last sentence "and we see from the definition that T is non-singular". Didn't we assume $T$ to be invertible a.k.a non-singular to begin with?
My next question is about the other half of the proof. This seems really confusing to me. First of all I don't understand why we again assume that $T$ is non-singular since we are trying to prove the backward direction now. Second of all, I don't understand why it follows from Corollary 9.7 that $T$ is a bijection. Corollary 9.7 simply gives an equivalence of several properties if $n=m$, but this does not mean that $T$ satisfies these properties, it only means that if it satisfies one of them, the others will follow automatically.
Can someone please clear things up for me? I've been trying to understand this proof for 3 days now, but it feels like I'm going in circles. Any help is appreciated. An alternative proof would also be helpful. Thanks very much!
 A: I agree that the theorem and its proof are poorly written.
Given a linear map $T\colon U\to V$ (were $U$ and $V$ are finite-dimensional vector spaces with dimensions $n$ and $m$ respectively) one can represent $T$ by a matrix $T_{\alpha}^{\beta}$ with respect to chosen bases $\alpha$ of $U$ and $\beta$ of $V$. Hence, after a choice of bases, one can talk about the matrix of a linear map. (Again, I stress that this matrix representation depends on a choice of bases!)
In the theorem above and the corollary, the author refers to both the linear map and a matrix representation as $T$. So this forces a lot of confusion into the situation.
I think the following is all you need to take away from this:

Theorem: Let $T\colon U\to V$ be a linear map. Then $T$ is invertible if and only if the conditions of corollary 9.7 hold. Moreover, if $T$ is invertible then $T_{\alpha}^{\beta}$ is a non-singular matrix for all bases $\alpha$ and $\beta$.

Proof: Assume that $T$ is invertible. Clearly $\ker(T)=\{0\}$ and $\text{im}(T)=V$. By theorem 9.6 we find that $n=m$. Moreover, as $T$ is assumed to be invertible, the conditions of corollary 9.7 hold. Conversely, assume that $n=m$ and that $T$ is bijective (in other words, assume the conditions of corollary 9.7), then there is nothing to prove.
Now let $\alpha,\alpha'$ be bases of $U$ and let $\beta,\beta'$ be bases of $V$. Note that $$T_{\alpha}^{\beta}=Id_{\beta'}^{\beta}T_{\alpha'}^{\beta'}Id_{\alpha}^{\alpha'}$$
and that the matrices $Q=Id_{\beta'}^{\beta}$ and $P=Id_{\alpha}^{\alpha'}$ are non-singular. It follows that $T_{\alpha}^{\beta}$ is non-singular if and only if $T_{\alpha'}^{\beta'}$ is non-singular.$\square$
Edit: Looking at the proof the author gave, it seems he interprets (v) of corollary 9.7 as saying that $T_{\alpha}^{\beta}$ is non-singular for some bases. From that he first wants to show that this then implies that $T$ is invertible as a linear map.
