In my Linear Algebra class I was introduced to the first isomorphism theorem between vector spaces in a way I didn't find anywhere else.
Quoting the statement:
Given two vector spaces $V,Z$ and a linear map $T\colon V\to Z$, there exists a unique injective linear map $\tilde{T}\colon V/\ker T\to Z$ such that $\tilde{T}(\pi(x))=T(x)$ (where $\pi\colon V\to V/\ker T$ is the canonical projection onto the quotient space).
$\tilde{T}$ is defined as $\ker T+x\mapsto T(x)$.
Now, I'm having trouble with proving the uniqueness of $\tilde{T}$.
In the proof we were given, there's the rather obscure statement «since it is well-defined, it is unique».
But how can it be?
I know that a linear map is unique given its action on a basis, but here we do not even mention a basis of $V/\ker T$.
So are there other methods to prove it?
Best Answer
The statement "since it is well-defined, it is unique" is certainly misleading. Uniqueness is shown (independently from existence = being well-defined) as follows: Assume $U\colon V/W\to Z$ is another linear map such that also $U(\pi(x))=T(x)$. Then $U=\tilde T$ because for any $y\in V/W$ there exists $x\in V$ such that $y=\pi(x)$ and then $U(y)=U(\pi(x))=T(x)=\tilde T(\pi(x))=\tilde T(y)$. Note specifically that we did not use any properties of $T$ for uniqueness, whereas well-definedness does require some additional properties ...