1-1 correspondence

Question

Let $A=\{1,2,3\}$ and $B=\{1,2\}$ and $f=\{(1,1),(2,2)\}$. Can we say that $f$ is a one-to-one mapping of $A$ onto $B$?

The thing here is that not every element of $A$ is "used" or mapped to $B$.

Answer 1

Preassuming that "mapping" is the same thing as function: no.

A mapping $f:A\to B$ is a subset of $A\times B$ with special property:$$\forall a\in A\exists!b\in B[\langle a,b\rangle\in f]$$

In words: for every $a\in A$ there exists exactly one $b\in B$ such that $\langle a,b\rangle$ is an element of $f$.

As you said yourself this is not satisfied in your case.

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In your case however $f$ is a set such that each of its elements is an ordered pair. That means that $f$ is indeed a mapping (or function, I would say). The domain of $f$ is then the set:$$\mathsf{Dom}(f):=\{a\mid \exists b[\langle a,b\rangle\in f]\}$$

So in your case $\mathsf{Dom}(f)=\{1,2\}$ and $f$ can be looked at as a function $\{1,2\}\to B$. This function is indeed one-to-one, because for that it is needed that for distinct elements $a,a'\in\mathsf{Dom}(f)$ the unique $b,b'\in B$ that satisfy $\langle a,b\rangle,\langle a',b'\rangle\in f$ are also distinct. This is satisfied in your case.

Answer 2

No, we can't. $f$ is a one-to-one mapping from $A\setminus\{3\}$ onto $B$. A function from something is always completely defined on the something unless you explicitly state that we are dealing with a partial function.