isomorphism of pointed sets What is an isomorphism in the category of pointed sets?
Is it just an exact sequence
$$ 1 \to A \to B \to 1 ?$$
(Note: even though the kernel of the middle map is zero, $A$ might not inject into $B$.)
I guess my confusion comes from the two (possibly distinct) notions of "bijection of pointed sets", which I interpret as a set-theoretic bijection preserving the distinguished point, and "isomorphism of pointed sets", which I'm not sure how to interpret.
If it helps, my motivation for this question comes from nonabelian cohomology (especially  $H^1$).
 A: In any category, an isomorphism of two objects $A,B$ is just a morphism $f$ from $A$ to $B$ so that there exists a two-sided inverse- a morphism $g:B\to A$ such that the composition in both directions is the identity.
Applying the definition to the category of pointed sets gives us that an isomorphism in the category of pointed sets between $(A,a)$ and $(B,b)$ where $A,B\in \textbf{Set}$ and $a\in A,b\in B$ is a map of sets $f:A\to B$ such that $f(a)=b$ and such that there exists a map $g:B\to A$ such that $g(b)=a$ and $f\circ g= id_B$, $g\circ f= id_A$. In particular, $f$ and $g$ are isomorphisms of sets, also known as bijections.
For better or worse, exact sequences $1\to A\to B\to 1$ in pointed sets are not in bijection with isomorphisms $A\to B$, unlike in module categories, where such a statement about sequences $0\to A\to B\to 0$ corresponding to isomorphisms $A\to B$ would be true. To see this, recall that an exact sequence of pointed sets means that the image of each map is the preimage of the distinguished point of the next map. Applying that to an exact sequence $$1\to A \to B \to 1$$ where $a\in A,b\in B$ are the distinguished points gives you that $A$ must surject onto $B$, but $A$ need not inject into $B$. Consider the following counterexample. Let $A=\{a,x,y\}$ with $a$ the distinguished point, and $B=\{b,z\}$ with $b$ the distinguished point. Let $f:A\to B$ be defined by $a\mapsto b$ and $x,y\mapsto z$. Then this fits in the exact sequence $1\to A\to B\to 1$ but is not a bijection.
