Module localization, Natural map, hom and bijectivity The following question is from my assignment on localization of modules and I need help with it.

Let V be an A-module, $S\subseteq A$ be a multiplicatively closed subset and let ${t_S}^V: V \to S^{-1} V, x\to x/1$ be a natural map.

(a) Prove that $\ker {t_S}^V=\{ x\in V | sx =0$ for some $s\in S\}$.
Unfortunately, I am not able to prove this and I think $\ker{t_S}^V$ should be all x in V such that x/1=0 ie x=0 because the map is $x\to x/1$. I don't understand why s should be there with x.
(b) The map ${t_S}^V$ is bijective iff all ${\lambda}_s $, $s\in S$ is bijective, where $\lambda_s : V \to V $  is the map $ v \to sv$.
Again the map : $x \to x/1$ is injective and surjective always.I am unable to correlate it with multiplication by s.
(c) The natural ring hom $t_S : A\to S^{-1}A$  is bijective iff $S\subseteq A^{\times}$.
Due to the problem in understanding the map, I am not able to attempt (c) also.
Please guide.
 A: For part (a): $x\in$ker$({t_S}^V)$ if and only if ${t_S}^V x=\frac x1=\frac 01$.  This is equivalent to there existing $s\in S$ such that $$sx=s(x\cdot1-0\cdot 1)=0.$$

If $\lambda_s$ is surjective for all $s\in S$, then given $s\in S$ and $v\in V$ we may find $x\in V$ with $sx=v$, so $t^V_S(x)=\frac x1=\frac vs$.  Thus $t^V_S$ is surjective.
You have already shown that if $\lambda_s$ is injective for all $s\in S$, then $t^V_S$ is injective.
Thus if $\lambda_s$ is bijective for all $s\in S$ then $t^V_S$ is bijective.
Conversely suppose $t_S^V$ is bijective. You have already shown that $\lambda_t$ is injective for all $t\in S$.
Given $v\in V$ and $s\in S$ we want to find $x\in V$ with $sx=v$, so we can deduce that $\lambda_s$ is surjective.
As we are given $t_S^V$ is surjective, we know there exists $x\in V$ such that $\frac x1=\frac vs$.  That is: $t(sx-v)=0$ for some $x\in V$ and $t\in S$.
As $\lambda_t$ is injective, we know $sx=v$.  Thus $\lambda_s$ is surjective, as required.

Part c is just the special case where $V=A$.  We know $t_S=T^A_S$ is bijective if and only if $\lambda_s$ is bijective for every $s\in S$.
If $\lambda_s$ is bijective then in particular $\lambda_s(y)=1$ for some $y\in A$.  Thus $sy=1$ and $s\in A^\times$.
Conversely, if $st=1$, then $\lambda_s\lambda_t=\lambda_t\lambda_s=1$, so $\lambda_s$ is bijective (it has a two-sided inverse $\lambda_t$).
So the $s$ for which $\lambda_s$ is bijective are precisely the elements of $A^\times$.
Thus $\lambda_s$ is bijective for every $s\in S$ precisely when $S\subseteq A^\times$.
