Exact time-dependent solutions of Maxwell's equations in Maxwell's fish eye show that perfect imaging is not an artifact of a drain at the image, although a drain is required for subwavelength resolution.PACS numbers: 42.30. Va, 77.84.Lf Perfect imaging with positive refraction [1][2][3] challenges some of the accepted wisdom of subwavelength imaging [4][5][6][7]. In particular, it requires a drain for perfect resolution [8]. Merlin [7] argues that the perfect focusing is an artifact of the drain. However, instead of discussing Maxwell's fish eye, he considers the focusing of light in a spherical mirror. The mirror serves as a simple model that resembles the fish eye, but this model is too simple: Maxwell's fish eye has imaging properties different from mirrors [9]. Furthermore, his reasoning is in conflict with experimental evidence [10]. Let us briefly explain how perfect imaging is achieved in Maxwell's fish eye and what the role of the drain is. Further details can be found in Ref. [9].In Ref.[2] we solved Maxwell's equations for electromagnetic radiation in Maxwell's fish eye (with ε = µ = n) by reducing the problem to the propagation of a scalar wave D; the electromagnetic field can be calculated from D by certain differentiations [11]. We thus only need to discuss the imaging properties of D, which does not constitute a simple model, but an exact representation of the electromagnetic Green tensor [2, 11] in Maxwell's fish eye.The fish eye is characterized in terms of a length R that defines the scale of the refractive index profile. For simplicity of notation, we measure space in units of R and time in units of R/c; in these units the speed of light in vacuum is 1. In our units the index profile of Maxwell's fish eye is given by n = 2 1 + r 2 .(The scalar wave D satisfies the equationwhere we include a source term on the right-hand side that corresponds to a point source at position r 0 acting in one moment of time that we set to t 0 = 0 without loss of generality. The wave equation thus describes a flash of light emitted at an arbitrary position and we can shift the time of emission to an arbitrary t 0 . Note that any light field can be thought of as a continuous superposition of such light flashes, and so our case is sufficiently general. Consider the Fourier transformation of the light flashNote that we can read the Fourier integral (3) as the amplitude of the wavethat is created by the continuous emission of light flashes at times t 0 with phases ωt 0 . The Fourier amplitude D thus plays a double role: it describes the frequency components of a single flash of light emitted at r 0 but also the amplitude of a stationary wave generated at the source point r 0 with frequency ω.To proceed, we write the wave equation (2) in the frequency domainthat has the solution [9]in terms of the Möbius-transformed radius [2, 11]We see that the source point r 0 corresponds to r ′ = 0. We also see that the image point of light rays in Maxwell's fish eye [11],corresponds to r ′ = ∞. The Möbius-transformed radius thus ...