I am not sure how much set theory you have done, but you probably do not understand 'injective function' or 'cardinality'. We say a function $f:A\to B$ is injective if and only if we have $a=b$ whenever $f(a)=f(b)$. Likewise, we say a function $f:A\to B$ is surjective if every member of $B$ can be written as $f(a)$ for some $a$ in $A$. We say a function is bijective if it is both surjective and injective, i.e. it is one-to-one.

Cardinality is a bit more subtle. We say two sets $A$, $B$ have the same cardinality if and only if there exists a bijective function from $A$ to $B$. Formally, the cardinality of a set $A$ is defined as a (proper) class of all sets which bijects with $A$.

Let $f:A\to B$, $g:B\to A$ be our injections. Now for each element $a$ in $A$, either it is $g(b)$ for some unique $b$, or it is not in the image of $g$. Similarly, every element $b$ of $B$ is either $f(a)$ for some unique $a$, or not in the image of $f$. Now we construct a backward chain for an element $a$ in $A$:-

$a$, $b$, $c$, $d$,...

in such a way that $a=g(b)$, $b=f(c)$, $c=g(d)$, chain may terminate or it may go on forever. Similarly, we construct the chain for each element of $B$. Given this chain, can we identify some bijection?

Yes, first, we note the set $A$ can be split into three disjoint subsets - those with infinite chain, those with finite odd-length chain (i.e. those terminates at some elements of $A$), and those with finite even-length chain. Similarly, $B$ can be split into three subsets. Now we see that an element $a$ of $A$ has odd (even) length chain if and only if $f(a)$ has even (odd) length chain, and similarly for $B$. Further, $f(a)(g(b))$ has infinite chain if and only if $a(b)$ has infinite chain. Hence we can form a one-one correspondence between those in $A$ with odd length chain with those in $B$ with even length chain by bijecting $f(a)$ with $a$, and bijecting $a$ with $b$ if $a$ has finite even length chain with $a=g(b)$, and for those with infinite chain, we can have a bijection $a$ with $f(a)$ (or $g(b)$ with $b$, just be consistent!).