If $T\circ S=id$ then $ker(S)=\{0\},ker(T)=\{0\}$ 
Let $T,S:V\to V$ then If $T\circ S=id$ then $ker(S)=\{0\}$and $ker(T)=\{0\}$

How can it be proven without the fact that for transformation to be bijection $S$ must be $1-1$ and $T$ onto
 A: Suppose $Sv=0$, then $TSv=0$ by linearity, but $TSv=v$ by hypothesis, so $v=0$. Thus $S$ has trivial kernel.
$T$ however need not have trivial kernel at the level of generality that you've posed the question. Indeed, although we have just shown that $S$ must be injective, if $S$ is not also surjective, then $T$ may fail to be injective and you may still have $T \circ S=id$. This can happen in infinite dimensional vector spaces: the key is that $T$ is injective when restricted to the range of $S$.
In finite dimensional vector spaces you have the invertible matrix theorem and the related rank-nullity theorem to tell you that this cannot happen.
A: It is not true in general vector spaces. Take the vector space with a countable basis $(e_1,e_2,\ldots)$ and make:
$$S(e_i)=e_{i+1}$$
$$T(e_i)=\begin{cases}0&i=1\\e_{i-1}&i\gt 1\end{cases}$$
Then $T\circ S=id$ but $\ker(T)\ne\{0\}$.
I can see you have other answers that prove $\ker(S)=\{0\}$ so I won't repeat it here.
A: If $Sv=0$, then with $T(S(v))=v$ we deduce that $T(0)=v$, but $T(0)=0$, so $v=0$, this proves $\ker(S)=\{0\}$.
Anyway, we have $\text{Im}(S)\cong V/\{0\}\cong V$, but it need no then $\text{Im}(S)=V$ unless $V$ is finite dimensional.
