The exponential map is not open on the von Neumann algebra of all bounded linear operators on an infinite dimensional separable complex Hilbert space. A 1987 article by Conway and Morrel shows that the spectrum of an element of the interior of the image of the exponential map does not separate 0 from infinity. On the other hand, a bilateral shift U has spectrum equal to the unit circle centered at the origin, and every Borel logarithm on the circle applied to the unitary operator $U$ yields a preimage point for U under the exponential map. Hence U is in the image but not in the interior of the image.

I learned about Conway and Morrel's article from this answer by David Speyer.

The exponential map is not open on the von Neumann algebra of all bounded linear operators on an infinite dimensional separable complex Hilbert space. A 1987 article by Conway and Morrel shows that the spectrum of an element of the interior of the image of the exponential map does not separate 0 from infinity. On the other hand, a bilateral shift U has spectrum equal to the unit circle centered at the origin, and every Borel logarithm on the circle applied to the unitary operator $U$ yields a preimage point for U under the exponential map. Hence U is in the image but not in the interior of the image.

I learned about Conway and Morrel's article from this answer by David Speyer.

The exponential map is not open on the von Neumann algebra of all bounded linear operators on an infinite dimensional separable complex Hilbert space. A 1987 article by Conway and Morrel shows that the spectrum of an element of the interior of the image of the exponential map does not separate 0 from infinity. On the other hand, a bilateral shift U has spectrum equal to the unit circle centered at the origin, and every Borel logarithm on the circle applied to the unitary operator $U$ yields a preimage point for U under the exponential map. Hence U is in the boundary of the image.

I learned about Conway and Morrel's article from this answer by David Speyer.

The exponential map is not open on the von Neumann algebra of all bounded linear operators on an infinite dimensional separable complex Hilbert space. A 1987 article by Conway and Morrel shows that the spectrum of an element of the interior of the image of the exponential map does not separate 0 from infinity. On the other hand, a bilateral shift U has spectrum equal to the unit circle centered at the origin, and every Borel logarithm on the circle applied to the unitary operator $U$ yields a preimage point for U under the exponential map. Hence U is in the image but not in the interior of the image.

I learned about Conway and Morrel's article from this answer by David Speyer.