Exercise 2.4.24

Answers

1.
If v + N(T) = v′ + N(T), we have that v − v′∈ N(T) and thus T(v) − T(v′) = T(v − v′) = 0.
2.
We have that
T¯((v + N(T)) + c(u + N(T))) = T¯((v + cu) + N(T))
= T(v + cu) = T(v) + cT(u).
3.
Since T is surjective, for all y ∈ Z we have y = T(x) for some x and hence y = T¯(x + N(T)). This means T¯ is also surjective. On the other hand, if T¯(x + N(T)) = T(x) = 0 then we have that x ∈ N(T) and hence x + N(T) = 0 + N(T). So T¯ is injective. With these argument T¯ is an isomorphism.
4.
For arbitrary x ∈ V , we have
T¯η(x) = T¯(x + N(T)) = T(x).
User profile picture
2011-06-27 00:00
Comments