For the following proposition, (taken from Arbib and Manes' Arrows, structures and functors text)
Proposition: Epi-mono factorizations are unique up to isomorphism in the sense that if $(e',m')$ is an epi-mono factorization
of a given map $f:A\rightarrow B$, then there is an isomorphism $\phi$ such that the above diagram commutes.
Proof: Let $c'\in C.$ Pick $a\in A$ with $e'(a)=c'.$ Define $\phi(c'))=e(a).$ Let $a\in A.$ Then there exists $a_1\in A$ with $e'(a)=e'(a_1)$ with $\phi(e'(a))=e(a_1).$ But $m(e(a))=f(a)=m'(e'(a))=m(e'(a_1))=m(e(a_1))$ and $m$ is one-to-one, so that $e(a)=e(a_1)=\phi(e'(a)).$ This proves that $\phi\circ e'=e$. It is then immediate that $\phi$ is onto since if $a\in A$ (so that $e(a)$ is a typical element of $f(A)$), $e(a)=\phi(c')$ for $c'=e'(a).$ $\color{red}{Let \, c'\in C'.\, Then \, \phi(c)=e(a)}.\,$ where $e'(a)=c'$ and so $m(\phi(c'))=m(e(a))=m'(e'(a))=m(c').$ This proves $m\circ \phi=m'$. Finally if $\color{blue}{c'\neq c''\in C'\, then\, since \,m'(c')\neq m'(c''), \,m(\phi(c'))\neq m(\phi(c''))\, so\, that\, surely \,\phi(c')\neq \phi(c'').}$
The relevant definition and theorem before the cited proposition are as follows:
Definition: We say that the pair $(e,m)$ epi-mono factorization of the map $f:A\rightarrow B$ if $e$ is an epimorphism and $m$ is a monomorphism such that $f=m\circ e$
Theorem: $f$ is a bijection iff $f$ is an isomorphism.
I am having a bit of trouble following the proof of the theorem. My difficulties are as follows:
$(1)$ The statement of the theorem implicitly assumes that the map $f$ already has a $(e,m)$ epi-factorization already and that existence of another $(e',m')$ is used to show that the map $\phi $ is an bijective and hence an isomorphism?
$(2)$ The highlighted red line of text, should it not read $\phi(c')=e(a)$, instead of $\phi(c)=e(a)?$
$(3)$ The highlighted blue line of text, I understand that the author is using the fact that if $x\neq y$, then $f(x)\neq f(y)$ which is the criteria for a function being one to one. So if $c'\neq c''\in C'$ then $m'(c')\neq m'(c'');$ with $m\circ \phi=m'$, so $m(\phi(c'))\neq m(\phi(c''))$. Because $m\circ \phi$ is assumed to being one to one, then we can conclude that $\phi(c')\neq \phi(c'')$ for the reason that $m\circ\phi$ is one to one means $\phi(c')\neq \phi(c'')$ then $m(\phi(c'))\neq m(\phi(c'')).$
Thank you in advance.
Best Answer
Since we are dealing with sets, we have the factorisation through the image $A\to f(A)\to B$, where we first restrict the codomain of $f$ from $B$ to its image, and then take the inclusion of the image into $B$. The first function is surjective (an epimorphism) and the second is injective (a monomorphism), so this is indeed an epi-mono factorisation! It's not guaranteed that these exist in all categories, as Mark Saving said. They do for sets and abelian categories though.
Now that we know an epi-mono factorisation exists, we can consider any such factorisations $A\to C\to B$ and $A\to C'\to B$. Because such factorisations exist, we may use them freely to obtain the isomorphism in the middle. This is similar to how products, coproducts etc. are shown to be unique up to isomorphism, so pls get used to this kind of arguments! :)
The red line is incorrect. It should be as you said, as the domain of $\phi$ is $C'$ and we want to send each $c'\in C'$ to $C$ and find some $a\in A$ which is sent to $\phi(c')$. What the author has accidentally written makes no sense! However, once fixed the way you did, this shows that $\phi$ is an epimorphism.
You have understood the blue part correctly yes. We have the assumption that $m$ and $m'$ are monomorphisms (injections) and that $m' = m\circ\phi$. It is a nice little result you can try to show as an exercise that the following holds: if $h=g\circ f$ and $h$ is a monomorphism (an epimorphism), then $f$ is a monomorphism ($g$ is an epimorphism). It follows also from this that $\phi$ must be a monomorphism.