Do strong epis have this property?

Question

I was reading some notes on category theory from Mariusz Wodzicki and exercise 51 says the following: Let $\epsilon$ be a strong epimorphism and $\mu$ a monomorphism such that $\mu \circ \epsilon$ is an epimorphism. Show that $\mu$ is an isomorphism. The author then proceeds to call these extremal epimorphisms in the next paragraph.

I don't see how the usual definition of extremal epis connects with this. In fact, I think the above is false: Unless I am mistaken, in the usual category of topological spaces Top, strong epis are precisely quotient maps. So if we take $\epsilon$ to be the quotient from an interval to $S^1$ then $\mu$ to be the identity from $S^1$ to itself (but in the target put the trivial topology for instance) then this seems to fail no?

Answer 1

Edited 2016-10-01 (Since I didn't read the question correctly)

You are correct. Strong doesn't imply this property.

Since identity morphisms are always strong this fact would mean that every monomorphism which is an epimorphism is an isomorphism. However, many such morphisms are known not to be isomorphisms. For example:

1. The inclusion of the natural numbers into the integers in the category of monoids.
2. The inclusion of the integers into the rationals in the category of rings.
3. The identity map from a non-empty set to itself considered as a morphism of topological spaces where the domain and codomain have the discrete and indiscrete topologies respectively (which is closely related to the second part of what you wrote).

Old answer (which assumed the usual definition of extremal)
Regarding the first part about strong and extremal .

A morphism $f:A\to B$ is called strong if for each commutative diagram $$\require{AMScd}\begin{CD} A @>{f}>> B\\ @V{u}VV @VV{v}V \\ S @>>{m}> C \end{CD}$$ where $m$ is a monomorphism, there exists a morphism $w: B\to S$ such that $mw=v$ (and hence $wf=u$). Now suppose that $f = qp$ where $q: T\to B$ is a monomorphism. It follows that the diagram $$\require{AMScd}\begin{CD} A @>{f}>> B\\ @V{p}VV @VV{1_B}V \\ T @>>{q}> B \end{CD}$$ commutes and hence there is a unique morphism $r: B\to T$ such that $qr=1_B$. This forces $q$ to be an isomorphism (since $q$ is a mono and split epi), proving that $f$ is extremal.

Regarding the second part. Let $\epsilon: I \to S_2$ be the quotient you mentioned. Now you seem to be suggesting that you want to $\mu : S_2 \to \tilde S_2$ to be the identity map but $\tilde S_2$ to have indiscrete topology. But then $\epsilon$ does not factor through $\mu$ (since they have different codomains).