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?



Notions of infinity in $\mathsf{ZF}$ without choice
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.




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You might be interested in the following paper:



> 
> Truss, J. **Classes of Dedekind finite cardinals.** *Fund. Math.* **84** (1974), no. 3, 187–208. [PDF](http://matwbn.icm.edu.pl/ksiazki/fm/fm84/fm84119.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.


