All we know is that there is a map from S to T; we call this $\alpha$. Likewise, all we know is that there is a map from T to S; we call this $\beta$. When we presuppose $\beta\alpha=1_S$, here's what we have: An $s$ maps, via $\alpha$, to some $t$ for all $s$ in $S$. These $t$s, __which are not necessarily are of $T$__ since we do not know if $|S|=|T|$, then map back to the original $s$ via $\beta$.
From this, we can conclude that my counter example holds perfectly for the _general_ notion that it is trying to disprove.