First isomorphism theorem in vector spaces: does "well-definedness" imply uniqueness? 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?
 A: 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 ...
A: Apart from the missing hypotheses, I think the following can be said about "since it is well-defined, it is unique". It is definitely not a very clear way of saying whatever is meant. I note first that in mathematical language uniqueness is not in general taken to imply existence: saying "an element with property $P$ is unique" means that at most one element $e$ makes $P(e)$ hold, but it does not claim such$~e$ exists. However changing to the definite article "the element with property $P$ is unique" does implicitly make the existence claim.
Now one is dealing here with a map $\tilde T$, which is specified by prescribing its values at arbitrary arguments. Since the prescription given gives at least one expression for the value at any given argument, such a map is automatically unique: there is no way that two different maps could both satisfy the prescription (and this is based merely on the form of the definition, not on any detailed arguments). However, the prescription actually gives more than one expression for the value at any given argument, since those arguments are classes, and the expression is in terms of a chosen member from that class. If two or more expressions for the value at the same argument would give different values, then the a map satisfying the prescription would not exist. This can be shown not to happen, and the conclusion is that the map is well-defined. So a more proper way to say this would be "it is a priori clear that such a map ($\tilde T$) is unique, and since it is well defined, it exists".
