Notions of infinity in $\mathsf{ZF}$ without choice Consider the following statements about a given set $X$ in in $\mathsf{ZF}$:
(1) There is $x_0\in X$ such that there is a surjective map $\varphi: X\setminus\{x_0\}\to X$.
(2) There is an injective map $\iota:\mathbb{N}\to X$.
It is easy to see that (2) implies (1) in $\mathsf{ZF}$, but are they equivalent?
 A: No, they are not equivalent.
It is a nice theorem that if there exists an infinite Dedekind-finite set (which is a set which satisfies the negation of (2)), then there is one which satisfies the first condition.
If $D$ is a Dedekind-finite set, then $S(D)$ which is the set of all injective finite sequences from $D$ is also Dedekind-finite (because the sets are injective, every collection of them is uniformly enumerated, so if there was a countable infinite set of these sequences, their union would be a union of uniformly enumerated sets, which would be a countably infinite subset of $D$).
Now simply consider the projection from $S(D)\setminus\{\varnothing\}$ onto $S(D)$ where you remove the last coordinate of the sequence.

You might be interested in the following paper:

Truss, J. Classes of Dedekind finite cardinals. Fund. Math. 84 (1974), no. 3, 187–208. PDF

In which the author takes seven definition of finiteness (proposed by Azriel Levy) and investigates the relations between them. Your first condition is one of the properties considered there.
