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 as 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.