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My textbook begins the chapter on equipotent sets like this: (English is not my first language so I'm sorry for poor choice of words/ terminology)

If we have $2$ sets $A$ and $B$ with a finite number of elements, it's theoretically easy to see whether they have the same number of elements. We do the following thing: we take one element of the set $A$ and one element of set $B$ and pair them. If we continue that process, after finitely many steps we will have one of the following situations:

$1.$ There are no unpaired elements in $A$ and no unpaired elements in $B$.

$2.$ The set $B$ doesn't contain any unpaired elements.

$3.$ The set $A$ doesn't contain any unpaired elements.

In the first case we constructed a bijection from set $A$ to $B$ and it's evident that $A$ and $B$ have the same number of elements.

In the second case, we constructed an injection from $B$ to $A$ and it's evident that $A$ has more elements than $B$.

In the third case, we constructed an injection from $A$ to $B$ and it's evident that $B$ has more elements than $A$.

Question: In the second case, why did we pick injection from $B$ to $A$? Wouldn't a function from $A$ to $B$ also be injective? And similarly for the third case, why not from $B$ to $A$?

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    In the second case, there are elements in $A$ which are not related to elements in $B$ (since all elements in $B$ are related to elements in $A$). By definition of a map/function, we only have a map from $B$ to $A$, not from $A$ to $B$ (recall that a map $f:X \to Y$ relates $\textbf{every}$ element of $X$ with some element in $Y$)! The same holds for the third case. – 2017-02-07
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    @Student Ohhhhhh, right....Thanks so much! – 2017-02-07
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    The 3 situations should be **mutually exclusive**, which is not the case. Situation 1 is a **special case** of situation 2, and also of situation 3. The conclusion about existence of injections is not harmed by that, because every bijection is also an injection. But the conclusions "$A$ has **more** elements than $B$" in 2, and "$B$ has more elements than $A$" in 3) are not okay. In the special case described as situation 1 both sets have the same number of elements. – 2017-02-07
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    E.g. situation 2 must be described as:"$B$ does not contain unpaired elements **and** we are not in situation 1". – 2017-02-07

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Let's see about case number 2. So basically you're pairing the elements of $A$ and $B$. Now, since the set $B$ contains no unpaired elements that would mean that $A$ has more elements than $B$. Basically in that case it makes more sense to make an injection from $B$ to $A$ to prove that, since in that case all elements from $B$ would be 'hit', but not all elements from A would, thus you conclude $B$ has less elements. If you would try to make an injection from A to $B$, from all elements of $A$ you would fail, since $A$ has more elements, and if $f: A \rightarrow B$ takes all elements of $A$, not all will have a unique element in $B$, thus it won't be an injection.

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    My question is silly... You can't make an injection from $A$ to $B$ since mapping $A\rightarrow B$ wouldn't be a function! – 2017-02-07
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    @Lewis If it's all clear now that's great! :) – 2017-02-07
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    Statement "$B$ contains no unpaired elements" does not allow us to conclude that "$A$ has more elements than $B$". This because the statement does not exclude that also $A$ contains no unpaired elements. – 2017-02-07
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    Pardon, my mistake, it is more or equal! I forgot to add that! And thank you! – 2017-02-07