Some $m\times n$ matrix $M$ has a singular value decomposition $M=U\Sigma V^H$, where $U,V$ are unitary and $\Sigma$ is a rectangular diagonal matrix with only real entries. $^H$ means conjugate transpose.
Let the column vectors of $U$ be denoted $u$ and called left-singular vectors, let the column vectors of $V$ be denoted $v$ and called right-singular vectors, and let $\sigma$ be a diagonal entry of $\Sigma$, called a singular value.
With all that out of the way, it is stated that $Mv=\sigma u$ and $M^Hu=\sigma v$. I understand these statements, and have proved them to myself as follows:
Let $e$ represent any unit vector that is a column of the identity matrix. Now,
$$Mv=U\Sigma V^Hv=U\Sigma e=U\sigma e=\sigma u; M^Hu=V\Sigma^TU^Hu=V\Sigma^Te=V\sigma e=\sigma v$$
The question:
Wikipedia states that a left singular vector corresponding to a singular value of zero lies in the kernel of $M$, i.e. $Mu=0$. This doesn't make sense to me, since $Mv=\sigma u=0$, but $Mu$ could be anything really - it doesn't simplify in the expansion as the above expressions do. They also say that a right singular vector with singular value zero is in the cokernel of $M$. This is a new term to me, but I assume it means the left-nullspace i.e. $vM=0$. Again, this makes no sense since it is $Mv$ that can equal zero, not $vM$.
Could anyone clear up "cokernel" for me, as well as clear up why Wikipedia said $Mu=0$, not $Mv=0$? I find it hard to believe that Wikipedia made a typo.
Many thanks.