  For cardinals $\mathfrak{a}$ and $\mathfrak{b}$, we write
$\mathfrak{a}=^\ast\mathfrak{b}$ if there are sets $A$ and $B$ of cardinalities
$\mathfrak{a}$ and $\mathfrak{b}$, respectively, such that there are partial
surjections from $A$ onto $B$ and from $B$ onto $A$. $=^\ast$-equivalence
classes are called surjective cardinals. In this article, we show that
$\mathsf{ZF}+\mathsf{DC}_\kappa$, where $\kappa$ is a fixed aleph, cannot prove
that surjective cardinals form a cardinal algebra, which gives a negative
solution to a question proposed by Truss [J. Truss, Ann. Pure Appl. Logic 27,
165--207 (1984)]. Nevertheless, we show that surjective cardinals form a
``surjective cardinal algebra'', whose postulates are almost the same with
those of a cardinal algebra, except that the refinement postulate is replaced
by the finite refinement postulate. This yields a smoother proof of the
cancellation law for surjective cardinals, which states that
$m\cdot\mathfrak{a}=^\ast m\cdot\mathfrak{b}$ implies
$\mathfrak{a}=^\ast\mathfrak{b}$ for all cardinals $\mathfrak{a},\mathfrak{b}$
and all nonzero natural numbers $m$.
