Gene asks: << What does "homotopy equivalent to S^n" mean? In general, two spaces X & Y are homotopy equivalent if there exist (continuous) maps f: X -> Y and g: Y -> X such that f o g is homotopic to the identity on Y and g o f is homotopic to the identity on X. But due a theorem of J.H.C. Whitehead and the simplicity of the homotopy groups of S^n (for any k with 0 < k < n, the continuous maps f: S^k -> S^n are all homotopic to each other, so the groups pi_k(S^n) for these k are all the 0 group), it readily follows that any connected n-manifold M with the same homotopy groups as S^n is automatically homotopy equivalent to S^n. In fact, to conclude that a connected n-manifold M is homotopy equivalent to S^n, it's sufficient -- by using Poincaré duality -- to assume only that pi_k(M) = 0 for k in the range 0 < k <= floor(n/2). * * * Dylan asks: << Does anybody know a minimal counterexample with compact manifolds? Not I. But I wonder if using (homeomorphic?) link complements in S^3 one could cook up an example by surgeries on the link, or gluing two of these complementds together in two different ways, so that non-homeomorphic 3-manifolds are obtained, but taking their product with the circle we get homeomorphic 4-manifolds. (A fertile place to look may be among pairs of non-homeomorphic but homotopy equivalent 3-manifolds.) --Dan