if $\ker(ST)=\{ 0 \}$, then ker {S)={0} and $\ker(T)=\{0\}$ 
Given two linear transformations from $V$ to $V$ such that $\ker(ST) = \{0\}$.
Is $\ker(S) = \{0\}$ and $\ker(T) = \{0\}$?

I know that if the composition of two functions is injective - it doesn't necessarily mean that both are injective, but I just couldn't find a counter example in the case of linear transformations :\
 A: As you already mentioned, the injectivity of $ST$ implies the injectivity of $T$. An injective linear map $V\to V$ is also surjective (assuming $V$ is finite-dimensional), that is, an isomorphism. By the same argument $ST$ is an isomorphism, and also $S=(ST)T^{-1}$.
If $V$ is not finite-dimensional you can easily find a counter example. Just take a sequence space $V$ and define $S$ and $T$ to be the left- and right-shift, respectively.
A: You're right that it isn't generally true of functions.  But linear transformations from a finite-dimensional linear space to itself are pretty restrictive, making them analogous to mappings from a finite set to itself in some ways.
The easiest way to prove it depends on how many tools you have on your belt.  If you can show that $L:\mathscr V\rightarrow\mathscr V$ with $\ker(L)=\{0\}$ implies that $L$ is invertible, then you can show that $ST$ is invertible, which definitely implies that $S$ and $T$ are both invertible.  An even shorter argument could talk about dimension of the range of the functions.
