## kernel of the adjoint operator and the cokernel of the operator

Let $L$ be a closed linear operator from Banach space $X$ to Banach space $Y$. Under the heading of Fredholm operators in the "Encyclopedic Dictionary of Mathematics" it says that if the range of $L$ is closed and the domain dense, then the dimension of the cokernel is equal to the dimension of the kernel of the adjoint of $L$. Unfortunately there is some ambiguity as to whether or not they are already supposing that the operator is Fredholm. So, my question is whether or not one needs to assume that $L$ is Fredholm in addition to the hypotheses stated above to conclude that the dimension of the cokernel is equal to the dimension of the kernel of the adjoint.

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I think you'd have more luck over at math.stackexchange.com A hint is that we can identify the dual of $Y/\operatorname(Im)(L)$ with $\operatorname(Im)(L)^\perp$ which is easily seen to be the kernel of $L^*$. So yes they have the same dimension. – Matthew Daws Nov 17 2011 at 7:54
To elaborate on Matthew's answer: If the kernel of the adjoint is finite-dimensional, then it is isomorphic to the cokernel of the operator you started out with. You do not need Fredholmness of $L$. You need the $L$ has a closed image and that the kernel of $L^*$ is finite-dimensional. – Orbicular Nov 17 2011 at 7:59