Relationship between mono-, epi- and isomorphisms in Category Theory This video defines mono- and epimorphisms this way:


*

*$f: a \to b$ is a monomorphism if $f \circ g = f \circ h \Longrightarrow g = h \; \forall g, h: c \to a$ (i.e. $f$ is left-cancelable)

*$f: a \to b$ is an epimorphism if $g \circ f = h \circ f \Longrightarrow g = h \; \forall g, h: b \to c$ (i.e. it is right-cancelable)


Q1: Is there a reason why the implications between equalities and equalities when composed with $f$ are not coimplications?
The video goes on saying that $f: a \to b$ is an isomorphism if $\exists f^{-1}: b \to a \;\mid\; f^{-1}\circ f = id_a \land f \circ f^{-1} = id_B$
Q2: Is there some relationship between $f$ being both mono- and epi- and it being an isomorphism?
One might naively go ahead and try to say for example that, if $f$ is an isomorphism
$$\forall g, h: c \to a, \quad f \circ g = f \circ h \Longrightarrow f^{-1} \circ f \circ g = f^{-1} \circ f \circ g \Longrightarrow id_A \circ g = id_A \circ h \Longrightarrow g = h$$
But step 1 assumes $f^{-1}$ to be left-cancelable, which is implied by $f$ being left-cancelable in the other sense of the implication, proof:
$$f \circ g = f \circ h \Longrightarrow g = h \\
\Downarrow \\
f \circ f^{-1} \circ g = f \circ f^{-1} \circ h \Longrightarrow f \circ g = f \circ h \\
\Downarrow \\
g = h \Longrightarrow f \circ g = f \circ h$$
So the above attempt to prove that $f$ is mono- if it is iso- depends on it being mono- (if the previous question has answer "no"). The epi- counterpart of this reasoning is trivial.
Consider that I managed to follow the video only up to the functors definition.
 A: Q1: the reverse implications are trivial.
Q2: an isomorphism is necessarily mono and epi. The converse is not true in general categories. For instance, the embedding of $\mathbb{Z}$ into $\mathbb{Q}$ is mono and epi in the category of rings, but it's clearly not an isomorphism.
On the positive side: in an abelian category, a morphism that's both mono and epi is an isomorphism.
Why is an isomorphism a mono? The idea is exactly the one you had, but inserting $f^{-1}$ in the right place: suppose $f$ is an isomorphism and that
$$
f \circ g = f \circ h
$$
Then also
$$
f^{-1}\circ(f \circ g) = f^{-1}\circ(f \circ h)
$$
and, by associativity,
$$
(f^{-1}\circ f) \circ g = (f^{-1}\circ f) \circ h
$$
giving
$$
\mathit{id}_a\circ g=\mathit{id}_a\circ h
$$
and, finally, $g=h$.
The proof that an isomorphism is epi is dual.
A: For $Q1$, the converse is avoided because it is trivially true for all $g,h$.
Q2: It is obviously true that every isomorphism is epi and mono, but the converse is not true. Take the category formed from a poset[*], where $\mathrm{Hom}(X,Y)$ always has at most one element. When $\mathrm{Hom}(X,Y)$ can never have more than one element, every morphism is epi and mono trivially, because every $g,h:c\to a$ are necessarily equal. However, in the category for a poset, the only morphisms with inverses are the indentities, where $X=Y$.
[*] Given any poset (partially ordered set,) $(P,\leq)$, you can define a category with objects $X\in P$, and with $$\mathrm{Hom}(X,Y)=\begin{cases}\{\star_{X,Y}\}&X\leq Y\\
\emptyset &\text{otherwise}
\end{cases}$$
with $\star_{Y,Z}\circ\star_{X,Y} = \star_{X,Z}$.
One of the conditions of a poset is that if $X\leq Y$ and $Y\leq X$ then $X=Y$. That shows that the only isomorphisms are the identities.
