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> Let $W$ be a finite-dimensional and $T\in \mathcal{L}(V,W)$. Show that if $T\in\mathcal{L}(V,W)$ injective then there exists $S \in \mathcal{L}(W,V)$ such that $ST$ is the identity map on $V$. I saw some posts on this statement but I didn't find the same approach as mine. If someone could give a feedback on my proof, I would really appreciate it. Thank you in advance.
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As $T\in \mathcal{L}(V,W)$ is injective, then by fundamental theorem of linear maps we have that $\dim V=\dim \text{null}\ T+\dim \text{range}\ T=0+\dim \text{range}\ T\le \dim W$. So, $\dim V\le \dim W$.

Let $(v\_1,...,v\_n)$ be a basis of $V$ and $(w\_1,...,w\_m)$ be a basis of $W$ (so $n\le m$). Define $T$ and $S$ as the following:

$T(v\_i)=w\_i, S(w\_i)=v\_i$ with $i=1,...,n$.

Both $T$ and $S$ are well defined as $n\le m$. Thus, $ST(v\_i)=S(w\_i)=v\_i$which is exactly an identity map as each $v\in V$ can be expressed uniquely.