91Ó°ÊÓ

A manifold \(M\) is called simply-connected if \(M\) is connected and if every smooth map \(f: S^{1} \rightarrow M\) is smoothly contractible to a point. [Actually, any space \(M\) (not necessarily a manifold) is called simply- connected if it is connected and any continuous \(f: S^{1} \rightarrow M\) is (continuously) contractible to a point. It is not hard to show that for a manifold we may insert "smooth" at both places.] (a) If \(M\) is smoothly contractible to a point, then \(M\) is simply-connected. (b) \(S^{1}\) is not simply-connected. (c) \(S^{n}\) is simply-connected for \(n>1\). Hint: Show that a smooth \(f: S^{1} \rightarrow S^{n}\) is not onto. (d) If \(M\) is simply-connected and \(p \in M\), then any smooth map \(f: S^{1} \rightarrow M\) is smoothly contractible to \(p\). (c) If \(M=U \cup V\) where \(U\) and \(V\) are simply-connected open subsets with \(U \cap V\) connected, then \(M\) is simply-connected. (This gives another proof that \(S^{n}\) is simply-connected for \(n>1\).) Hint: Given \(f: S^{\prime} \rightarrow M\), partition \(S^{1}\) into a finite number of intervals cach of which is taken into either \(U\) or \(V\). (f) If \(M\) is simply-connected, then \(H^{1}(M)=0 .\) (See Problem 7.)