Invertible Linear Map Suppose $V$ is finite-dimensional and $S,T\in \mathcal{L}(V)$. Prove that if $ST$ is invertible, then both $S$ and $T$ are invertible. 
This is the partial question retrieved from Linear Algebra Done Right, and I have come out with different solution from the solution guide. 
My solution: 
Since $ST$ is invertible, then the only possibility for $(ST)u=0$ is when $u=0$. Therefore I have $(ST)(0)=0=S(T(0))=0$. Since $T(0)=0$ (as proven in book), I must have $S(0)=0$. Since $T(0)=0$ and $S(0)=0$ comes directly from $(ST)(0)=0$, both $S$ and $T$ are injective. Since both of them are linear operator, we can deduce they are both invertible (by equivalece statement.)
Is my proof valid?

As José Carlos Santos stated, your proof is not correct. You have to show two things: injectivity and surjectivity of both $S$ and $T$. If you want to use zero element in your proof then injectivity can be shown by proving the following implication $$ f(u) = 0 \Rightarrow u = 0 \ ,$$ 
that is, you do not assume in the beginning that $u$ is zero. 
Other hint that might be useful in the proof: If $ST$ is injective then $T$ must be injective. If $ST$ is surjective then $S$ must be surjective.   
