Ensuring that a function is well defined.

Question

From my prior studies that a relation between two sets $R\subseteq A\times B$ is deemed a function if and only if it satisfies the following properties.

Definition-1

• $\forall a \in A \, \exists b\in B \, ((a,b)\in R)$

• $\forall a \in A \, \forall b_1,b_2 \in B \, ((a,b_1)\in R\land(a,b_2)\in R\implies b_1=b_2)$

But while attempting the following problem $T$ is injective if and only if $ST=I_V$ for some $S\in\mathcal{L}(W,V)$. I seemed to have proved incorrectly that the function in this problem is well defined, so much so that i know feel that i have completing misunderstood the concept.

For example the fellow that corrected me seemed to be saying that to prove that a function is well defined should prove the following.

Definition-2

• $\forall a,a'\in A(a=a'\implies f(a) = f(a)')$.

  I know that the second clause in the definition one shows that a function is well defined, so how does the second clause in the definition two communicate the same idea?

  I hate to grovel but i really need help with this seeing as how this is such a fundamental concept and i didnt even realize that had it wrong this whole time!

Answer 1

What you call Definition-2 is not a definition. Obviously, it assumes that the expression $f(a)$ as already been defined.

When we try to define a function $f\colon A\longrightarrow B$, there are two things that may go wrong:

1. The definition may be such that $f(a)$ is actually not defined for some $a\in A$. That's the case, for instance, if you talk about $f\colon\mathbb{R}\longrightarrow\mathbb R$ defines by $f(x)=\frac1{x^2-1}$.
2. An element $a\in A$ may have more than one image. That happens, for instance, if you define $f\colon\mathbb{Q}\longrightarrow\mathbb Z$ as $f(q)=\text{denominator of }q$.

Checking that a function is well-defined means chacking that these problems don't occur.

Answer 2

When you want to define a function $f:A\rightarrow B$ by means of an assignment or a rule $x\mapsto y$, it is your duty to show that $y$ is determined unequivocally by $x$, which generally ammounts to demonstrating that the assignment is independent of the representation of $x$.

Answer 3

You misunderstand the comment to your other question. You have written

Assume that $S(Tv+u)=w_1$ and $S(Tv+u)=w_2,$ then by definition of $S$ we have $Tv=w_1$ and $Tv=w_2$ then given the injectivity of $T$ we have $w_1=w_2$

As @Suzet noted, this is tautological. If we have, $$w_1=S(Tu+v)=w_2,$$ then of course $w_1=w_2,$ provided that $S(Tu+v)$ means something, that is, provided it is well-defined.

To see that $S$ is well-defined, you have to show either that there is only one way of writing any element $x$ of the domain as $Tu+v$ or that all such ways lead to the same value of $S$.

For example, suppose we try to define a function $f$ on the positive integers by saying for every $n$ write n=$a+b$ where $a$ and $b$ are positive integers, and define $f(n)=ab.$ This is not well-defined, because we can choose different values of $a$ and $b$ that lead to different values of $f(n).$