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.

## Best Answer

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.