Let $V$ be a vector space (with respect to some scalar filed $\mathcal{F}$), and let $M$ be a linear vector space of $V$. Consider the equivalence relation $\sim$ such that for any $x,y\in V$, $x\sim y$ if $y-x\in M$ and let $x+M$ be the coset of $x$ modulo $M$ and $V/M = \{x+M:x\in V\}$ be the coset of $V$ modulo $M$. Let also $Q:V\to V/M$ denote the quotient mapping. It is possible to make $V/M$ a vector space, called the quotient vector space, in which $Q$ is a linear mapping with $Ker\,Q = M$ (see Theorem 2.6.1 of Berberian (1992). Linear Algebra. OUP). Thus, so far we got that $Ker\,Q$ equals the "zero vector" of $V/M$.
We know that for any linear mapping $T:V\to W$, with $W$ a vector space, $T$ is injective iff $Ket,T=\{\theta\}$, where $\theta$ is the zero vector of $W$ (as well as V). Why this result does not apply to $Q$? Is it because $V$ and $V/M$ have different structures (V is a set, whereas V/M is a collection of sets)?
A linear mapping $T:V\to W$ is injective iff $\operatorname{Ker}T=\{0\}$, with $0$ the zero vector of $V$, not of $W$. In the case of quotient mapping, its kernel is the subspace $M$, which in general is not the trivial subspace $\{0\}$. $Q$ would be injective if $0$ would be the only vector of $V$ which is mapped to the zero vector of the quotient space, that is to the coset $M$, but instead, you have that all the vectors of $M$ are mapped to that coset, not only $0$.