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Let $T\colon X\to Y$ be a bounded linear operator acting between Banach spaces. Suppose $T$ is an isomorphism onto its range. Must $T^{**}\colon X^{**}\to Y^{**}$ be an isomorphism onto its range also?

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By the closed range theorem, an operator $T$ has closed range if and only if the range of its adjoint is closed. Using this fact twice we see that $T$ has closed range if and only if $T^{\ast\ast}$ has closed range.

On the other hand, since $T$ is injective with closed range, we have that $T^\ast$ is onto, hence $T^{\ast\ast}$ is injective, too.

Combining these two observations gives that $T^{\ast\ast}$ is an isomorphism onto its range.

See Chapter 4, Theorems 4.12 and 4.14, in Rudin's book on functional analysis for proofs of the assertions used here.

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