# [Math] Relationship between mono-, epi- and isomorphisms in Category Theory

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.

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.
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, $g=h$.