Let $V$ and $W$ be subspaces, assume $\dim(V)=\dim(W)$, and that $T:V\rightarrow W$ and $U: W \rightarrow V$ be linear and satisfy $U*T=\mathrm{id}_v$ or $TU=\mathrm{id}_w$, then T is an isomorphism. (id stands for identity mapping)
Proof: Assume $U*T=\mathrm{id}_v$, so $UT$ is both injective and surjective,hence $T$ is injective. $\tag1$ By rank-nullity theorem, $\mathrm{rank}(T)$= $\dim(V)- \dim(N(T))=\dim(W)$, hence $T$ is surjective. So $T$ is an isomorphism.
My question is how does it deduce that $T$ is injective in ($1$)?
Let $u$ and $v$ two vectors in $\textsf{V}$ such that $\textsf{T}(u) = \textsf{T}(v)$. We need to show that $u=v$ in order to prove the injectivity of $\textsf{T}$, right? But this follows inmediatelly from the fact that $\textsf{UT} = \textrm{id}_\textsf{V}$, see: \begin{align} u &= \textrm{id}_\textsf{V}(v) \\ &= \textsf{U}(\textsf{T}(v)) \\ &= \textsf{U}(\textsf{T}(w)) \\ &= \textrm{id}_\textsf{V}(w) \\ &= w. \end{align}