Several years ago [8] we proved that Hubert space is homeomorphic with both its unit sphere {#: ||x|| = 1} and its unit cell {x:||x||^l}. Later [9] we showed that in every infinite-dimensional normed linear space, the unit sphere is homeomorphic with a (closed) hyperplane and the unit cell with a closed halfspace. It seems probable that every infinite-dimensional normed linear space is homeomorphic with both its unit sphere and its unit cell, but the question is unsettled even for Banach spaces. Corson [4] has recently proved that every fc$ 0 -dimensional normed linear space is homeomorphic with its unit cell. In the present note, we establish the same result for a class of infinite-dimensional Banach spaces which is believed to include all such spaces. It is proved to include every infinite-dimensional Banach space which is reflexive, or admits an unconditional basis, or is a separable conjugate space, or is a space CM of all bounded continuous real-valued functions on a metric space M.We employ the following tools:(1) If E and F are Banach spaces and u is a continuous linear transformation of E onto F, then there exist a constant ra£]0, <» [ and continuous mapping v of F into E such that uvx = x, vrx = rvx, and \\vx\\ rSw||#|| for all x£F and rÇiR (the real number space). If G is the kernel of u and hy= (uy, vuy-y)(E:FXG for each yÇzE, then h is a homeomorphism of E onto FXG. Let \\(p, q)\\ =max (\\p\\, \\q\\) for all (p, q)GFXG, and let £y = (|M|/||ft:y||)A:y for all y£E. Then £ is a homeomorphism of E onto FXG which carries the unit cell of E onto that of FXG.(2) If S is a closed linear subspace of a Banach space £, then E is homeomorphic with the product space (E/S) XS and the unit cell of E is homeomorphic with the unit cell of this product space (with respect to any norm compatible with the product topology).(3) In each infinite-dimensional normed linear space, the unit cell is homeomorphic with a closed halfspace.(4) If Q is an open halfspace in an infinite-dimensional normed linear space and p is a point in the boundary of Q, then QVJ {p} is homeomorphic with Q.