# Second Dual of Kernel of a Surjective Bounded Linear Operator Between Banach spaces.

Let $$X$$ and $$Y$$ be Banach spaces and let $$T : X \to Y$$ be a bounded linear operator. Suppose that $$T$$ is surjective, and thus by the Open Mapping Theorem, $$T$$ is open.

I have come across a paper that then $$(\ker T)^{**}$$ and $$\ker (T^{**})$$ are isomorphic. I can't seem to see why. Perhaps I am missing a theorem?

Any help would be appreciated.

Here's my thoughts so far. Since $$T$$ is surjective:

$$X/\ker (T) \cong Y$$

Is it true that $$T^{**}$$ is also surjective? If so, then:

$$X^{**}/\ker(T^{**}) \cong Y^{**}$$

Thus:

$$\left ( X / \ker (T) \right )^{**} \cong X^{**} / \ker (T^{**})$$

Is this enough to conclude that $$(\ker (T))^{**} \cong \ker (T^{**})$$? And if so, I do not see where the openness of $$T$$ is used.

• Maybe the closed range theorem helps. – gerw Apr 10 at 7:20

## 1 Answer

I suggest to write your problem in terms of (topologically) short exact sequences (SES) in the category of Banach spaces: For $$K=$$ker$$(T)$$ you then have a SES $$0\to K \stackrel{i}{\to} X\stackrel{T}{\to} Y\to 0.$$ The dual sequence $$0\to Y^* \stackrel{T^*}{\to} X^*\stackrel{i^*}{\to} K^*\to 0$$ is again exact (the exactness at $$X^*$$ requires ker$$(i^*)=$$im$$(T^*)$$ and follows from the the fact that $$T$$ is open and the exactness at $$K^*$$ is the Hahn-Banach theorem, that $$T^*$$ and $$i^*$$ are open onto their images follows from the open mapping theorem). Going to the second duals we get the SES $$0\to K^{**} \stackrel{i^{**}}{\to} X^{**}\stackrel{T^{**}}{\to} Y^{**}\to 0.$$ Hence, $$K^{**}$$ is isomorphic to ker$$(T^{**})$$. Moreover, they are not just isomorphic (by some isomorphism constructed in a fancy way) but canonically isomorphic (the bitransposed of the inclusion $$i$$ is an isomorphism).