HW 9
1. Show that if a:G×M→M is proper, then θ=(πM,a):G×M→M×M is proper. But converse is not true.
Take a compact set K⊂M×M, then θ−1(K)∈a−1(π2(K)), where π2:M×M→M is the projection to the second factor. Since π2 is continuous, π2(K) is compact; since a is proper, a−1(π2(K)) is compact. θ−1(K) is closed subset of a compact set, hence is compact.
For a counter-example of the converse statement, consider G×G→G×G, where G=R and action is addition.