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**1. The problem statement, all variables and given/known data**

Let T: V [tex]\rightarrow[/tex] Z be a linear transformation of a vector space V onto a vector space Z. Define the mapping

[tex]\bar{T}[/tex]: V/N(T) [tex]\rightarrow[/tex] Z by [tex]\bar{T}[/tex](v + N(T)) = T(v)

for any coset v+N(T) in V/N(T).

a) Prove that [tex]\bar{T}[/tex] is well-defined; that is, prove that if v+N(T)=v'+N(T), then T(v)=T(v').

b) Prove that [tex]\bar{T}[/tex] is linear.

c) Prove that [tex]\bar{T}[/tex] is an isomorphism.

**2. Relevant equations**

Proving part (a) will prove that the transformation is a one-to-one transformation. This is important for part (c).

In order to prove part (b), use the fact that for T: V[tex]\rightarrow[/tex]W, T is linear if and only if T(cx + y) = cT(x)+T(y) for all x,y [tex]\in[/tex] V and c [tex]\in[/tex] F.

To prove that a transformation is an isomorphism, part (c), there must exist a linear transformation [tex]\bar{T}[/tex] that is invertible ([tex]\bar{T}[/tex][tex]^{-1}[/tex]). [tex]\bar{T}[/tex][tex]^{-1}[/tex] exists iff [tex]\bar{T}[/tex] is one-to-one and onto.

**3. The attempt at a solution**

(a) If v+N(T) = v'+N(T), then v+N(T)-N(T) = v'+N(T)-N(T). Thus v = v'. The vectors v and v' are the same without the nullspace of T.

If v=v', then T(v) = T(v'). This also means it is one-to-one.

I'm not sure if what I have is a valid proof for part (a). The part that I'm not certain about is if I am actually able to subtract the nullspace of T from both sides.

(b) If [tex]\bar{T}[/tex] = T(v), let cx+y = v+N(T). Then [tex]\bar{T}[/tex](cx+y) = T(cx+y-N(T)) = cT(x)+T(y)-T(N(T)). This is the first part of the proof.

I am having trouble with the next part of the proof. I have to prove that c[tex]\bar{T}[/tex](x) + [tex]\bar{T}[/tex](y) = cT(x)+T(y)-T(N(T)) (the right hand side is from the first part of the proof), but I'm not sure how to plug x and y into [tex]\bar{T}[/tex].

(c) A transformation is an isomorphism if it is one-to-one and onto.

To prove one-to-one: Since v+N(T)=v'+N(T), then v=v'. If v=v', then T(v)=T(v'). T(v)=T(v') implies that v=v', this is the definition of one-to-one.

I'm having trouble proving that [tex]\bar{T}[/tex] is onto. From the question, we know that T is a linear transformation that is onto, but how do I relate it to [tex]\bar{T}[/tex]?

Any and all help is much appreciated!