Let $T : V → W$ be a linear transformation of vector spaces. We say that
$T$ is invertible if and only if there exists a map $S : W → V$ such that
$S \circ T = 1_W$ and $T \circ S = 1_V$ . Show that $T$ is an isomorphism if and only
if $T$ is invertible.
My thoughts on the problem is as follows: Since I know we call $2$ vector spaces isomorphic if and only if there exists linear maps $α: V → W$ and $β: W → V$ such that $α \circ β = \text{Id}_W$ and $β \circ α = \text{Id}_V$. Thus if $T$ is invertible if and only if there exists a map $S : W → V$ such that $S \circ T = 1_W$ and $T \circ S = 1_V$, then it should suffice to say that $T$ is an isomorphism if and only if $T$ is invertible.
Although I believe my logic is correct, I am sure this not an acceptable proof. I was just looking for any insight on how I could possibly improve my proof so that it would become acceptable.
Thanks for any help!
Best Answer
Let $V$ and $W$ be two vector spaces and let $T$ be a Linear Transformation. T is said to be an isomorphism if T is also a Bijection. Bijection implies that there exists another Linear Transformation $S:W\to V$ such that S and T are each other's inverse.
So essentially the statements that T is an Isomorphism and T is Invertible are one and the same. Let me give an outline of the proof
Let T be an isomorphism. To prove T is invertible, we need to show that there exists a linear transformation $S:W\to V$ that maps each element of W uniquely onto V. Thus T is invertible. Let $y\in W$. Since T is onto, there exists an $x\in V$ such that $T(x) = y$. Uniqueness of x is established by T being one-to-one. So, $\forall y\in W, \exists$ unique $x \in V$ and hence define $$S:W\to V \quad \text{such that} \quad S(y) = x$$ where $T(x) = y$. Checking for Linearity of S is simple.
For the other way proof, using the two linear maps that compose to give identity, we can prove that T is ono-to-one and onto and therefore an isomorphism.