The following is from the paper "A Group Epimorphism is Surjective" by C.E. Linderholm
The proof of the proposition is the contrapositive of the statement from this post which have I have questions about.
Background information
Let $\phi: F\rightarrow G$ to mean that $F$ and $G$ are groups and $\phi$ is a homomorphism defined on $F$ and taking values in $G$. If $\psi:G\rightarrow H$ the composite homomorphism from $F$ to $H$ is written $\phi\psi.$ If $x\in F$ the image under $\phi$ is written $(x)\phi.$ A homomorphism $\psi$ that cancels on the right in every equation $\phi_1\psi=\phi_2\psi$ is a monomorphism. If $\psi:G\rightarrow H$ then $\psi$ is a monomorphism if for each group $F$ and for each pair $\phi_1, \phi_2$ of homomorphism from $F$ to $G$ such that $\phi_1\psi=\phi_2\psi,$ the equation $\phi_1=\phi_2$ holds. A homomorphism $\phi$ that cancels on the left in every equation $\phi_1\psi=\phi_2\psi$ is an epimorphism. If $\phi:F\rightarrow G$ and if an inverse homomorphism $\psi$ from $G$ to $F$ exists - i.e., one such that $(x)\phi\psi=x$ for each element $x$ of $F$ and $(y)\psi\phi=y$ for each element $y$ of $G$- then $\phi$ is an isomorphism. We prove that "A Group Epimorphism is Surjective".
Proof: Let $\phi$ be an epimorphism from a group $G$ to a group $H,$ and let $A$ be the image subgroup. We must show that $A=H.$ To do this we construct two homomorphism $\psi_1,\psi_2$ from $H$ to a group $K$ such that $\phi\psi_1=\phi\psi_2,$ and we use the resulting equation $\psi_1=\psi_2$ to prove that $A=H.$
Let $H/A$ be the set of all right cosets of $A;$ let $A'$ be something that is not a right coset of $A,$ and let $S$ be the set $H/A\cup\{A'\}.$ Let $K$ be the group of all permutations (bijections) of $S.$ If $h,h_1,h_2\in H$ and if $Ah_1$ and $Ah_2$ are the same coset, then $A(h_1h)$ and $A(h_2h)$ are the same coset, and hence the function from $A/H$ to $A/H$ that sends $Ah'$ to $A(h'h)$ is well defined. It is easily seen to be bijective. (in fact its inverse is given by $Ah'\rightarrow Ah'h^{-1}.$) If $A'$ is sent to itself, this defines a bijection of $S,$ which we write $(h)\psi_1.$ The function $\psi_1$ from $H$ to $K$ so defined is easily seen to be a homomorphism. Let $\sigma$ be the permutation of $S$ that interchanges $A$ and $A'$ and leaves everything else fixed, and if $h\in H$ write $(h)\psi_2$ for the composite permutation $\sigma^{-1}((h)\psi_1)\sigma.$ Then the function $\psi_2$ is the composition of $\psi_1$ with an inner automorphism of $K,$ and hence is a homomorphism from $H$ to $K$.
If $a\in A,$ then $(a)\psi_1$ leaves both $A$ and $A'$ fixed, and $\sigma$ leaves every other element of $S$ fixed Hence $\sigma$ and $(a)\psi_1$ commute, and $(a)\psi_1=\sigma^{-1}((a)\psi_1)\sigma.$ Since $\psi_1$ and $\psi_2$ agree on the range of $\phi,$ the equation $\phi\psi_1=\phi\psi_2$ holds. Since $\phi$ is an epimorphism, $\psi_1=\psi_2.$ Hence $(h)\psi_1$ commutes with $\sigma$ for each element $h$ of $H.$ Since $(h)\psi_1$ leaves $A'$ fixed and $\sigma$ interchanges $A'$ and $A.$ It follows that $(h)\psi_1$ leave $A$ fixed. Since $(h)\psi_1$ sends $A$ to $Ah,$ it follows that $h\in A.$ Since $h$ was an arbitrary element of the group $H,$ it follows that $A=H$ and $\phi$ is surjective.
Note: where it says the function from $A/H$ to $A/H$, should it not be $H/A$ to $H/A?$
Questions
There are some points about the above proof I am not clear on.
I understand that $A=\phi(G)$, but do the set $K$ consist of both the permutations $\sigma$ and $(h)\psi_1.$
For the function $(h)\psi_1$, it is defined as $(h)\psi_1:H\to K$ but what about at various places where it says it sends $A\to Ah,$ $A'$ to itself. In terms of elements evaluated by $\psi_1$, what is the difference between when it sends $A'$ to itself versus $(a)\psi_1.$ Earlier in the proof, it already states that $(h)\psi_1$ sends $A'$ to itself, but $(a)\psi_1$ leaves both $A$ and $A'$ fixed. The notation for the elements of $\psi_1$ are confusing. $h, a$ are elements of $H, A$ respectively, then when it says $(h)\psi_1$ leaves $A'$ to $A'$, what about $(h)\psi_1$ where it sends $A$ to $Ah.$ Also in the case of $A$ to $Ah$, does it mean $a\mapsto ah?$ Other points of confusion for me is how do both $\sigma, (a)\psi_1$ and $\sigma, (h)\psi_1$ commute, lastly, how does $(a)\psi_2=\sigma^{-1}((a)\psi_1)\sigma=(a)\sigma_1.$ Thank you in advance.
