# Suppose that $dim(U) = dim (V)<\infty$ and let $T\in Hom(U,V)$. If $AT=\iota_V$ or $TA=\iota_U$, then $A$ is an isomorphism and $A^{-1}=T$.

I have a question regarding the proof of this corollary 2.11, which I have encountered in a course on Linear Algebra. As this is a corollary, I have a feeling that the proof should be somewhat obvious and self-explanatory but I fail to see it.

Theorem 2.10

Assume that both $$U$$ and $$V$$ both have finite dimension $$n$$. If $$A$$ is surjective or injective, then $$A$$ is bijective.

Corollary 2.11

Suppose that $$dim(U) = dim (V)<\infty$$ and let $$T\in Hom(U,V)$$. If $$AT=\iota_V$$ or $$TA=\iota_U$$, then $$A$$ is an isomorphism and $$A^{-1}=T$$.

Proof. If $$AT=\iota_V$$, then $$A$$ is surjective, and if $$TA=\iota_U$$ then $$A$$ is injective. In both cases Theorem 2.10 implies $$A$$ is bijective. Then $$A^{-1}=T$$ follows.

What I do not understand is why $$A$$ is surjective when we have $$A\circ A^{-1}$$ but it is injective when $$A^{-1}\circ A$$.

If $$AT=\iota_V$$, then, for each $$u\in U$$, $$A(T(u))=u$$. Therefore, $$A$$ is surjective.
And, if $$TA=\iota_U$$, then, if $$v_1,v_2\in V$$ are such that $$A(v_1)=A(v_2)$$, then $$T(A(v_1))=T(A(v_2))$$, which means that $$v_1=v_2$$. So, $$A$$ is injective.