Does Surjective-infinite implies Dedekind infinite Suppose we define a set $A$ to be Surjective-infinite iff there is an surjection $f:A \rightarrow A$ such that it is not an injection. Is this true that $A$ is also dedekind-infinite, i.e there exist a $f^*$ from $A$ to $A$ that is injective but not surjective. 
Conceptually, using AC if need be. I could use the given $f$ to construct $f^*$ by sort of 'removing' the duplicates from the pre-image of $a$ under $f$, meaning if $f(c)=f(b)=a,$ define $f^*(a)$ to just be $b$.
If this is true, how do I do it more concretely ?
Cheers
 A: Assuming the axiom of choice, yes, quite easily. If $f\colon A\to A$ is a surjection, there is some $g\colon A\to A$ which is an injection and $f(g(a))=a$. Simply prove that since $f$ is not injective, $g$ cannot be surjective.
Not assuming choice, however, this is not necessarily true. It is consistent that there is a Dedekind-finite set $A$ and $f\colon A\to A$ which is surjective and not injective. For example, if $A$ is an infinite Dedekind-finite set, the set $S(A)$ of all finite injective sequences from $A$ is also Dedekind-finite and the function $f\colon S(A)\to S(A)$ given by erasing the last coordinate of a sequence is surjective but far from injective.
A: As stated in @AsafKaragila's answer, assuming the axiom of choice yes, it is true. Here's an explicit construction:
Define in $A$ the equivalence relation:
$$a\sim a' \stackrel{(def.)}{\iff} f(a)=f(a')$$
Denote with $[a]$ the equivalence class of $a$, and define the map $\psi_f\colon (A/\sim)\longrightarrow A$ via $\psi_f([a]):=f(a)$. The map $\psi_f$ is well-defined and bijective:

*

*(good-definitedness) $a'\in [a] \Longrightarrow [a']= [a] \Longrightarrow \psi_f([a'])=\psi_f([a])=f(a)$;

*(surjectivity) by the surjectivity of $f$: $\forall b\in A, \exists a\in A\mid b=f(a)=\psi_f([a])$;

*(injectivity) $\psi_f([a])=\psi_f([a'])\Longrightarrow f(a)=f(a')\Longrightarrow a\sim a'\Longrightarrow [a]=[a']$.

Moreover, by the axiom of choice, there's a bijection $\varphi\colon (A/\sim)\longrightarrow R$, where $R\subsetneq A$ is a complete set of representatives (the inclusion is strict, because $f$ is not injective). Therefore, $\phi\colon R\longrightarrow A$ defined by $\phi:=\psi_f\circ \varphi^{-1}$ is a bijection (as composition of bijections); so, $\phi^{-1}\colon A\longrightarrow R$ is a bijection. Finally, $f^*\colon A\longrightarrow A$ defined by $f^*(a):=\phi^{-1}(a)$ is a injection but not a surjection because $f^*(A)=\phi^{-1}(A)=R\subsetneq A$.
