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?

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    $\begingroup$ The obvious question: What does "abstract-algebraic isomorphism" mean for you? $\endgroup$ – Lord_Farin Apr 12 '13 at 23:03
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    $\begingroup$ There is only one notion of isomorphism ( = invertible morphism), used throughout in mathematics. $\endgroup$ – Martin Brandenburg Apr 12 '13 at 23:06
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    $\begingroup$ Except for what they teach first-year undergraduates (bijective homomorphism)... $\endgroup$ – Zhen Lin Apr 12 '13 at 23:08
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    $\begingroup$ @MartinBrandenburg It's not the correct general meaning, but it is correct for most of the early algebraic examples, so it is often the first definition given. It is correct there. The fact that you need another definition when you want to be more general is the point of this question, I think. $\endgroup$ – Thomas Andrews Apr 12 '13 at 23:20
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    $\begingroup$ Defining an isomorphism as a bijective homomorphism is as morally wrong as defining compactness as being closed and bounded. $\endgroup$ – Michael Greinecker Apr 12 '13 at 23:51

Notice that Awodey writes "Moreover, in some cases only the abstract definition makes sense, for example, in the case of a monoid", not "in the case of monoids".

For Awodey, a monoid is a category with one object, and the isomorphisms are the invertible arrows, i.e. the invertible elements of the monoid. Since the arrows aren't actually functions, it doesn't make sense to ask if they are bijective.

  • $\begingroup$ Aha, excellent catch Alex! $\endgroup$ – Zev Chonoles Apr 12 '13 at 23:51

Note that he does not say

more general than the one in abstract algebra


... has the advantage over other possible definitions ...

As is discussed in the comments under your question, it is often the case that beginning abstract algebra students are taught that the meaning of "isomorphism" (of groups, rings, etc.) is "bijective homomorphism". That is false. It is a property that happens to characterize the isomorphisms of those particular objects, but that does not make it the definition. Some would argue that pedagogical concerns make this an acceptable falsity that, one hopes, will be corrected later.

Edit: Alex Kruckman has noticed what I think Awodey meant by his comment about a monoid. Though I think what I said below may still be helpful to you, so I'll leave it.

Recall that a monoid is a set $S$ with an associative binary operation $\star$ for which there is an identity element $e_S\in S$ (necessarily unique). Given two monoids $(S,\,\star,\,e_S)$ and $(T,\,\bullet,\,e_T)$, a monoid homomorphism from $S$ to $T$ is a function $f:S\to T$ such that

  • $f(s_1\mathbin{\star}s_2)=f(s_1)\mathbin{\bullet} f(s_2)$ for all $s_1,s_2\in S$

  • $f(e_S)=e_T$

Such homomorphisms can be bijective; thus, there is a notion of bijective monoid homomorphism.


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.


Another example : there exists continuous bijective functions that are not homeomorphisms.


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