Q.If V is an arbitrary vector space over $F$ and if $T$ belongs $Hom(V,V)$ is right-invertible with a unique right inverse, prove that its invertible. This was question asked in Topics in Algebra by I.N. Herstein. I was able to prove this result by using the argument by contradiction, let us assume $T$ is not invertible and So there exist a $S\not=0$ such that $ST=TS=0$ and let $S_1$ is unique right inverse, now $ S_1TS=IS=S=0$, which is a contradiction. I had question if the vector space is not finite dimensional does the statement holds true? If yes, does my argument above holds.
In general the statement is not true. The problem is that $S_1T$ is the identity on the image of $T$ which could be smaller than $V$. Think for example in the space of polynomial ring the multiplication by $x$ which is linear and has an inverse in the image (i.e. divide by $x$ by shifting all the coefficient) but clearly is not invertible on the all space (a polynomial of degree zero has not pre image)