Linear transformation is injective if and only if there exists a linear transformation where the composition is identity How to prove that if $U$ and $V$ are finite dimensional vector spaces and $T:U \rightarrow V$ is a linear transformation, then $T$ is injective if and only if $S \circ T$ is the identity function on $U$ for some linear transformation $S$?
 A: Suppose $U$ and $V$ are finite-dimensional vector spaces and $T:U \rightarrow V$ is a linear transformation. Define $R: range(T) \rightarrow V$ with $R(T(u))=u$; this can be done because $T$ is injective, that is, for each $T(u) \in range(T)$ for some $u \in U$, we have $T^{-1}(T(u))=\{u\}$, a singleton. Now, for $u,u' \in U$, and $\alpha$ a scalar, $$R[\alpha T(u)+T(u')]=R[T(\alpha u+u')]=\alpha u+u'=\alpha R(T(u))+R(T(u'));$$ $R$ is then a linear transformation and there exists an extension of $R$, a linear transformation $S:V \rightarrow U$ that satisfies $(ST)(u)=S(T(u))=R(T(u))=u$ whenever $u \in U$. Therefore $S \circ T$ is the identity on $U$. Conversely, suppose there exists a linear transformation $S:V \rightarrow U$, which satisfies $(ST)(u)=u$ whenever $u \in U$. If $T(u) \in range(T)$ for some $u \in U$, and if $u' \in T^{-1}(T(u))$, then $u'=S(T(u'))=S(T(u))=u$. Therefore $T$ is injective.
A: First off, $S\circ T=I_U$ clearly implies (requires) that $T$ is injective; no linear algebra is involved here.
Remember that to define a linear map (here $S$), you can pick a basis $\mathcal E$ of its domain space ($V$) and send each of its vectors to arbitrarily chosen vectors in the codomain ($U$). This is the fundamental fact behind the of representation of linear maps by matrices: the coordinates on some basis $\mathcal B$ of those vectors of the codomain define the columns of the defined linear map (with respect to the bases $\mathcal E,\mathcal B$).
Now let $\mathcal B=[b_1,\ldots,b_n]$ be a basis of $U$. Since $T$ is injective, their images $[T(b_1),\ldots,T(b_n)]$ are a linearly independent family in $V$. Therefore set $e_i=T(b_i)$ for $i=1,\ldots,n$; this need not be a basis of $V$, but being independent can always be extended to a basis $[e_1,\ldots,e_m]$, with $m\geq n$. Now we want $S\circ T=I_U$, so we should define $S(e_i)=b_i$ for $i=1,\ldots,n$. This does not completely define $S$, since we must also specify $S(e_i)$ for $n<i\leq m$; however this can be done in an arbitrary way (for instance choosing all those $S(e_i)$ to be the zero vector). The fact that $S(T(b_i))=b_i$ for all $b_i$ in the basis $\mathcal B$ ensures by linearity that $S\circ T=I_U$, as desired.
