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Let $\mathcal{A}:X\to Y$ be continuous linear operator, $X$ and $Y$ are Banach spaces. Let $\text{Im} \mathcal{A}=Y$.

Is $\ker\mathcal{A}$ a complemented subspace of $X$?

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No, not in general. For example, if $X$ is not isomorphic to a Hilbert space then $X$ contains a noncomplemented subspace $Z$. The quotient map $X \longrightarrow X/Z$ is surjective, continuous and linear, but the kernel (which is $Z$ of course) is not complemented.

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    I would like to point you to another question which is related to this one: http://math.stackexchange.com/questions/2119896/can-every-closed-subspace-be-realized-as-kernel-of-a-bounded-linear-operator-fro If the restriction of this question to the case that $X=Y$ would result in a positive answer, this would result in a negative answer to the question I have linked above. @Norbert2017-02-01