1. let $K$ be a category with equalizers, show that every extremal epimorphism is epic.

  2. for composable morphisms $f: A \rightarrow B $ and $g: B \rightarrow C$ in $K$, show that if $gf$ is an extremal epimorphism, then $g$ is an extremal epimorphism.

  3. let $K$ be a category with pullbacks. For composable morphisms $f$ and $g$ as given in #2, show that if $g$ and $f$ are extremal epimorphisms, then $gf$ is an extremal epimorphism.

  4. show that if an extremal epimorphism is monic, then it is an isomorphism.

here is what I know:

In any category $K$, an arrow $f: A \rightarrow B$

  • is called an isomorphism if there is an arrow $g: B \rightarrow A$ in $K$ such that $ g \circ f = I_{A} $ and $ f \circ g = I_{B}$

  • is called a monomorphism, if given any $g,h : C \rightarrow A, fg=fh$ implies $g=h$

  • is called an epimorphism, if given any $i,j: B \rightarrow D, if = jf$ implies $i=j$

  • is an extremal epimorphism, if for each commutative diagram, $f = mh$ where if $m$ is monic, then $m$ is an isomorphism

I have tried to answer #2:

  • since $gf$ is an extremal epimorphism, then $gf = mh$ where $m$ is monic and an isomorphism.

  • now let $g = nk$ where $n$ is monic.

I now need to show that $n$ is an isomorphism.

my problem is how do I do this? I'm trying to visualise things using diagrams, but I'm lost.

Here is what I have so far for #1:

let $f: A \rightarrow B$ be and extremal epimorphism.

consider $ g,h: B \rightarrow C$ such that $gf=hf$ so I need to prove $g=h$

let $e: E \rightarrow B$ be the equaliser of $g, h$

now I don't know where to go from here to prove $g=h$.

I know that $f$ factors through the equaliser.

since $e$ is an equaliser, there is a unique morphism

$\varphi : A \rightarrow E$ therefore I now have $f=e \varphi$

which means that $e$ is now isomorphic since $f$ is an extremal epimorphism.

So can I now conclude $g=h$ ?

  • $\begingroup$ Do you only want the answer for #2? $\endgroup$ – Najib Idrissi Dec 3 '13 at 21:54
  • $\begingroup$ @nik no, I'd like hints to get all the answers. I had been able to work through #2, so I put what I had up. $\endgroup$ – sarah jamal Dec 3 '13 at 22:04
  • $\begingroup$ I expanded my answer to give hints for the other questions. $\endgroup$ – Najib Idrissi Dec 3 '13 at 22:22
  • $\begingroup$ @nik I have put up my attempt to answer #1, I am getting stuck though $\endgroup$ – sarah jamal Dec 10 '13 at 7:48

For #2: Suppose $g = mh$ where $m$ is monic, then $gf = (mh)f = m(hf)$, $m$ is still monic, and therefore since $gf$ is extremal epic $m$ is an iso. Therefore $g$ is extremal epic.

Basically you have to go back to the definition every time. At each step there is only one thing you can do, so you do it.

  • You want to prove that $g$ is extremal epic, so you write it as $g = mh$ with $m$ monic.
  • You know something about $gf$, so you compose the previous equation with $f$.
  • Et caetera.

Hints for the other ones:

  • #1: Let $f : A \to B$ be extremal epic. You want to prove that it is epic, so take any $g,h : B \to C$ such that $gf = hf$. You know that $K$ has all equalizers, so take the equalizer of $g$ and $h$. You know $gf = hf$, so by definition $f$ factors through the equalizer. The map from the equalizer to the domain is monic...

  • #3: Suppose $gf = mh$ where $m$ is monic. Write this equality in the form of a commutative square. Use the universal property of pullbacks. A good lemma to prove is that a pullback of a monomorphism is a monomorphism...

  • #4: Suppose $f : A \to B$ is extremal epic and monic, then write $f = f \circ \mathrm{id}_A$...

  • $\begingroup$ can I answer #3 like this: let $g = nk$ where $n$ is monic and an isomorphism. therefore $gf = nkf = n(kf)$ since $n$ is monic an isomorphism, $gf$ is an extremal epimorphism. $\endgroup$ – sarah jamal Dec 3 '13 at 22:23
  • $\begingroup$ No, you cannot. You need to consider every possible decomposition of $gf$ into $mz$ with $m$ monic, not just the ones of the form $mkf$. $\endgroup$ – Najib Idrissi Dec 3 '13 at 22:25
  • $\begingroup$ for #3, how do I use the property of pullbacks? To write $gf$ in form of a commutative square, does that mean $fg = nkmh$ where $f = mh$ with $m$ monic and ismorphism? $\endgroup$ – sarah jamal Dec 3 '13 at 22:43
  • $\begingroup$ still on #3, do i consider a pullback of $f$ and $m$? and then a pullback of $g$ and $n$? if so, where do I go from here? $\endgroup$ – sarah jamal Dec 4 '13 at 9:30
  • $\begingroup$ for question 1 can I then say the equaliser is isomorphic? and therefore g=h? $\endgroup$ – sarah jamal Dec 10 '13 at 21:14

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