I'm looking at a textbook exercise that asks to prove or give a counterexample that $\forall A : A \leq_m A \cup \{0\}$.$A \leq_m B$ is defined as:$\exists f$ total, computable s.t. $\forall x \in A : x \in A \Leftrightarrow f(x) \in B$,It then says in the answer sheet something to the effect of:The only counterexample is $A = \mathbb{N} \setminus \{ 0 \}$.Otherwise there is always an appropriate $f$:If $0 \in A$, $id$ is good enoughIf $0 \not\in A$, one such $f$ is$f'(x) = \begin{cases} x_0 & x = 0\\ x & otherwise\end{cases} $  for some $x_0 \not\in A \cup \{0\}$.However, before checking the answer sheet I was persuaded there were multiple such counterexamples.In fact, I still don't understand what's wrong in my proof that $A = \{ 1\}$ is a counterexample.It goes like this:Let $A = \{1\}$.  Then $A \cup \{0\} = \{0,1\}$.   Suppose such an $f$ as required by the definition exists.There are three cases, mutually exclusive by definition of  function:$(1,0) \in f$, therefore $f(A) = \{0\} \neq A \cup \{0\} = \{0,1\}$$(1,1) \in f$, therefore $f(A) = \{1\} \neq A \cup \{0\} = \{0,1\}$$(1,c) \in f$ for some other $c$, therefore $f(A) = \{c\} \neq A \cup \{0\} = \{0,1\}$Therefore $f$ such that $f(A) = A \cup \{0\}$ cannot exist for $A = \{ 1 \}$. QED.In particular, I don't see how, for the given$f'(x) = \begin{cases} x_0 & x = 0\\ x & otherwise\end{cases} $it holds that $f'(\{1\}) = \{1,0\}$It seems to me that $f'(\{1\}) = \{1\} \neq \{0,1\}$.What am I misunderstanding about either basic set theory or the definition of $\leq_m$?
Many-one reducibility: why $A = \{1\}$ is not a counterexample to $\forall A : A \leq_m A \cup \{0\}$?
computability;reductions;sets