If T1 :Rn→Rm and T2 :Rm→Rk are linear transformations, and if T1 is not onto, then neither is T2 ◦ T1.
I think it's False b/c if T2 (ran(T1)) = Rk then T2◦T1 is onto? Is there a more generic way for me to say that? THANKS!
Yes, the claim is false, but you need a more specific counterexample, not a more general one. Let n = k = 1 and m = 2, let T1 be an inclusion of R^1 into R^2 and T2 be a projection of R^2 onto R^1. Then T2 ◦ T1 is the identity, which is onto.
For arbitrary (not necessarily linear) functions, if T2 ◦ T1 is onto than so is T2, and if T2 ◦ T1 is an injection, then so is T1. One should be skeptical about stronger claims.