Informal writeup of the proof that group homomorphisms are epi iff su…

…rjective (#726)

This PR only adds an informal writeup of a proof that a group
homomorphism is surjective iff it is an epi. A proof without excluded
middle can be given using the theory of acyclic maps, which I'm still
working on with Ulrik and Tom. I thought that the writeup could still be
worthwhile to have on our website.

---------

Co-authored-by: Fredrik Bakke <fredrbak@gmail.com>

src/group-theory/epimorphisms-groups.lagda.md

@@ -22,10 +22,17 @@ open import group-theory.precategory-of-groups
 
 ## Idea
 
-A group homomorphism `f : x → y` is a epimorphism if whenever we have two group
-homomorphisms `g h : y → w` such that `g ∘ f = h ∘ f`, then in fact `g = h`. The
-way to state this in Homotopy Type Theory is to say that precomposition by `f`
-is an embedding.
+A [group homomorphism](group-theory.homomorphisms-groups.md) `f : G → H` is an
+**epimorphism** if the precomposition function
+
+```text
+  - ∘ f : hom-Group H K → hom-Group G K
+```
+
+is an [embedding](foundation.embeddings.md) for any
+[group](group-theory.groups.md) `K`. In other words, `f` is an epimorphism if
+for any two group homomorphisms `g h : H → K` we have that `g ∘ f = h ∘ f`
+implies `g = h`.
 
 ## Definition
 
@@ -61,3 +68,44 @@ module _
   is-epi-iso-Group =
     is-epi-iso-Large-Precategory Group-Large-Precategory l3 G H f
 ```
+
+### A group homomorphism is surjective if and only if it is an epimorphism
+
+**Proof using the law of excluded middle:** The forward direction of this claim
+is the easier of the two directions, and this part of the proof doesn't require
+the [law of excluded middle](foundation.law-of-excluded-middle.md). If `f` is
+[surjective](foundation.surjective-maps.md) and `g h : H → K` are two group
+homomorphisms such that `g ∘ f ＝ h ∘ f`, then to show that `g ＝ h` it suffices
+to show that `g y ＝ h y` for any `y : H`. Since we are proving a
+[proposition](foundation.propositions.md) and `f` is assumed to be surjective,
+we may assume `x : G` equipped with an
+[identification](foundation.identity-types.md) `f x ＝ y`. It therefore suffices
+to show that `g (f x) ＝ h (f x)`, which was assumed.
+
+For the converse, suppose that `f : G → H` is an epimorphism and consider the
+[image subgroup](group-theory.images-of-group-homomorphisms.md) `I := im f` of
+`H`. We first show that `I` is [normal](group-theory.normal-subgroups.md), and
+then we show that `I ＝ H`.
+
+In order to show that `I` is normal, we want to show that `I` has only one
+conjugacy class, namely itself. Consider the group `K` of permutations of the
+set of [conjugate](group-theory.conjugation.md)
+[subgroups](group-theory.subgroups.md) of the subgroup `I` of `H`. There is a
+group homomorphism `α : H → K` given by `h ↦ J ↦ hJh⁻¹`, where `J` ranges over
+the conjugacy classes of `I`. Notice that `I` itself is a fixed point of the
+conjugation operation `J ↦ f(x)Jf(x)⁻¹`, i.e., `I` is a fixed point of
+`α(f(x))`. We claim that there is another homomorphism `β : H → K` given by
+`h ↦ α(h) ∘ (I h⁻¹Ih)`, where we precompose with the
+[transposition](finite-group-theory.transpositions.md) `(I h⁻¹Ih)`. This
+transposition is defined using the law of excluded middle. However, note that
+`I` is always a fixed point of `β(h)`, for any `h : H`. Furthermore, we have
+`α(f(x)) ＝ β(f(x))`. Therefore it follows from the assumption that `f` is an
+epimorphism that `α ＝ β`. In other words, `I` is a fixed point of any
+conjugation operation `J ↦ hJh⁻¹`. We conclude that `I` is normal.
+
+Since `I` is normal, we may consider the
+[quotient group](group-theory.quotient-groups.md) `H/I`. Now we observe that the
+quotient map maps `f(x)` to the unit of `H/I`. Using the assumption that `f` is
+an epimorphism once more, we conclude that the quotient map `H → H/I` is the
+[trivial homomorphism](group-theory.trivial-group-homomorphisms.md). Therefore
+it follows that `I ＝ H`. This completes the proof.