Problem: Let $ A, B $ be two sets s.t. $ A \sim B $. Given set $ X $ denote $ H(X) = \{ f: X \rightarrow X | f \text{ IS NOT INJECTIVE} \} $. Show $ | H(A) |=| H(B) |$.
Note: I'm not allowed to use Cantor-Schreder-Bernstein theorem. So far in the exercises in the course we mainly focused on showing explicit bijections from one set to the other to prove they have same cardinalities, so that's the approach I was trying to take in my proof.
What I tried to do but couldn't: I was trying to define a bijection as follows " Let $ F: H(A) \rightarrow H(B) $ be defined as follows, $ (F(f))(b) = \_ \_ \_ ~, ~ \forall f\in H(A),\forall b \in B $. " and then I'd proceed to show that indeed it is a bijection. I somehow have to use the bijection from the given $ A \sim B $ and I need to make $ F(f) $ be non-injective.
I spent alot of time trying to make progress but it was of no avail, help would be very appreciated.
Let $g:B\to A$ be any bijection, and define $F(f):H(A)\to H(B),f\mapsto g^{-1}\circ f\circ g$.
We need to show that $F(f)\in H(B)$ i.e. $F(f)(x)$ is not one-one for $f:A\to A$ not one-one. Suppose both $a_1,a_2(\ne a_1)\in A$ map to $a\in A$ under $f$. Let $a_1=g(b_1),a_2=g(b_2)$. Clearly $a_1\ne a_2\implies b_1\ne b_2$ so$$F(f)(b_1)=g^{-1}(f(g(b_1)))=g^{-1}(f(a_1))=g^{-1}(a)\\F(f)(b_2)=g^{-1}(f(g(b_2)))=g^{-1}(f(a_2))=g^{-1}(a)$$which is the same as $F(f)(b_1)$. This gives that $F(f)(x)$ is a many-one function.
Now to show that the function $F$ is bijective:
$(1)$ One-one: $F(f_1)=F(f_2)\implies g^{-1}\circ f_1\circ g=g^{-1}\circ f_2\circ g$. Thus $g\circ(g^{-1}\circ f_1\circ g)\circ g^{-1}=g\circ(g^{-1}\circ f_2\circ g)\circ g^{-1}$ giving $f_1=f_2$.
$(2)$ Onto: Given $f_1\in H(B),g\circ f_1\circ g^{-1}\in H(A)$ and maps to $f_1$ under $F$.