questions about extremal epimorphisms in category theory 
*

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

*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.

*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.

*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 $g=h$ ?
 A: 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 $f = f \circ \mathrm{id}_A$...
