We first prove as a lemma that for any $B\subset A$, if there is an injection $f:A\to B$, then there is also a bijection $h:A\to B$.

Inductively define a sequence $(C_{n})$ of subsets of $A$ by $C_{0}=A\setminus B$ and $C_{{n+1}}=f(C_{n})$. Now let $C=\bigcup_{{k=0}}^{\infty}C_{k}$, and define $h:A\rightarrow B$ by

If $z\in C$, then $h(z)=f(z)\in B$. But if $z\notin C$, then $z\in B$, and so $h(z)\in B$. Hence $h$ is well-defined; $h$ is injective by construction. Let $b\in B$. If $b\notin C$, then $h(b)=b$. Otherwise, $b\in C_{k}=f(C_{{k-1}})$ for some $k>0$, and so there is some $a\in C_{{k-1}}$ such that $h(a)=f(a)=b$. Thus $h$ is bijective; in particular, if $B=A$, then $h$ is simply the identity map on $A$.

To prove the theorem, suppose $f:S\to T$ and $g:T\to S$ are injective. Then the composition $gf:S\to g(T)$ is also injective. By the lemma, there is a bijection $h^{{\prime}}:S\to g(T)$. The injectivity of $g$ implies that $g^{{-1}}:g(T)\to T$ exists and is bijective. Define $h:S\to T$ by $h(z)=g^{{-1}}h^{{\prime}}(z)$; this map is a bijection, and so $S$ and $T$ have the same cardinality.