Let's say we have the category of sets, and say we have a morphism between two objects in this category. Then, this morphism is in itself a set, since functions are defined through sets. So, let's say we have two functions (between two fixed objects $f,g:A \to B$) which are isomorphic in the category when considered as objects themself (sets), would this imply they are isomoprhic as objects in the arrow category as well? Would this always be the case even in general categories? If not, what constraints are needed so that it is?
If two morphisms are isomorphic as sets in a category, then are they isomorphic as objects in the arrow category
category-theory
Best Answer
The answer is no.
Consider a function $f : A \rightarrow B$. It is an arrow in the category of sets, but as you mention, it is a set too: namely, $f \subset A \times B$. So, we can consider $f$ to be an object in the category of sets. Your questions asks whether, given two functions $f,g : A \rightarrow B$, an isomorphism $f \cong g$ as objects in $Set$ implies an isomorphism $f \cong g$ as objects in the arrow category.
To show why this is false, consider the two simple facts.
A function $f : A \rightarrow B$ has cardinality $|A|$. Indeed, $f$ can be written as $\{(a,f(a)) : a \in A\}$, so the map $A \rightarrow f : a \mapsto (a,f(a))$ is a bijection.
If $f$ is an isomorphism in a category, and $f \cong g$ in the arrow category, then $g$ is also an isomorphism. Suppose $f : A \rightarrow B$ and $g : C \rightarrow D$. If $f$ is an isomorphism, it has an inverse $f^{-1}$, and if $f \cong g$, then there are invertible maps $u : A \rightarrow C$ and $v : B \rightarrow D$ such that $g \circ u = v \circ f$. Then $g \circ (u \circ f^{-1} \circ v^{-1}) = Id_D$ and $(u \circ f^{-1} \circ v^{-1}) \circ g = Id_C$, so $g$ is invertible too.
Now, a counter-example is easy. Take $f(x)=x$ and $g(x)=0$ as functions $\mathbb{N} \rightarrow \mathbb{N}$. They have the same domain, hence the same cardinality, so they are isomorphic as sets. However, they cannot be isomorphic in the arrow category, since $f$ is an isomorphism and $g$ is not.