TITLE: Why is 'isomorphism' defined more generally in Category theory than in Abstract Algebra?
QUESTION [8 upvotes]: I read in Awodey's Category Theory book that the definition of isomorphism in category theory is more general than the one in abstract algebra. For example, he says, the definition of isomorphism from abstract algebra doesn't make sense for monoids. Why not?

REPLY [4 votes]: A morphism of posets $f : (P,\leq) \to (Q,\leq)$ is a map $f : P \to Q$ with $x \leq y \Rightarrow f(x) \leq f(y)$. A bijective morphism of posets is a morphism such that the underlying map is bijective. In order to be an isomorphism, $f$ has to satisfy the stronger condition $x \leq y \Leftrightarrow f(x) \leq f(y)$. For example, $(P,\Delta_P) \to (P,P \times P)$ is a bijective homomorphism, but not an isomorphism (unless $|P| \leq 1$).
This is quite similar to the category of topological spaces. A bijective continuous map $f$ doesn't have to be an isomorphism (usually called homeomorphism in this category). It is an isomorphism iff the converse to continuity, namely openness, also holds: $U$ is open when $f^{-1}(U)$ is open.
But for algebraic structures it turns out that bijective morphisms are already isomorphisms (since one can check that the inverse map is also compatible with the operations).
In general, a (forgetful) functor $U : C \to D$ is called conservative if the following property holds: If $f$ is a morphism in $C$ and $U(f)$ is an isomorphism, then $f$ is an isomorphism. The above discussion provides examples of conservative and non-conservative forgetful functors.