I have been commanded on homework to find a non-bijective isomorphism in a category whose objects are sets, whose morphisms are set maps, and composition is the usual function composition. So our category is a subcategory of the category of ((sets)).

But, I think that such an isomorphism is in fact impossible to find.


Let $\mathcal{C}$ be a category of sets with morphisms being set maps and composition being the usual set composition. If $\mathcal{C}$ is the empty category, we have no isomorphisms, as we have no morphisms. Thus $\mathcal{C}$ contains at least a single object.

Take $A,B \in \mathcal{C}$ and assume we have an isomorphism $f \in Mor(A,B)$.

(Note that $A=B$ is allowed).

Since $f$ is an isomorphism, there exists $g \in Mor(B,A)$ such that $f \circ g = 1_B$ and $g \circ f = 1_A$. We show that $f$ must in fact be a bijection.

Injective: Take $a_1$ and $a_2$ in $A$ and suppose that $f(a_1) = f(a_2)$. We then have the following chain of equality: $$(g\circ f)(a_1) = 1_A(a_1) = a_1 = a_2 = 1_A(a_2) = (g\circ f)(a_2).$$

Hence we have that whenever $f(a_1) = f(a_2)$, we have $a_1 = a_2$. So $f$ must be injective.

Surjective: Take $b$ in $B$. We know that the morphism $g$ takes elements of $B$ to $A$. Hence $g(b) \in A$. We then have $$f(g(b)) = (f\circ g)(b) = 1_B(b) = b.$$

Thus $f$ maps $g(b) \in A$ to $b$ and $f$ is therefore surjective.

We can therefore conclude that any isomorphism in $\mathcal{C}$ must be bijective.

  • 1
    $\begingroup$ Indeed, any isomorphism in any subcategory of $\mathbf{Set}$ must be an isomorphism in $\mathbf{Set}$ itself, i.e. bijective. Are you sure you have not misread the question? $\endgroup$ – Zhen Lin Sep 5 '14 at 18:42
  • 1
    $\begingroup$ The professor hints that we need not assume that the identity morphism is what we usually take as the identity morphism. So perhaps the category in question is not actually a subcategory of ((sets)). $\endgroup$ – Vladhagen Sep 5 '14 at 18:45

Given the hint that the identity morphisms can be weird in ((sets)), we can proceed as follows. Let ((sets)) consist of just the one set $\{0,1\}$ and the one map, from $\{0,1\}$ to itself that sends both 0 and 1 to 0. This one map is then the identity map, composition is as usual, and this identity map is a non-bijective isomorphism.


I believe that this is what we need to examine. Although we intuitively want to assume that the identity morphism is the usual "identity" map, this is actually not stated as necessary in the problem.

Let $\mathcal{C}$ be a category consisting of a single set $A=\{1,2\}$ as its object, morphisms as set maps, and composition as usual.

Let $Mor(A,A)$ consist of the single morphism $f$ defined as follows:

$$f:\begin{cases}1 \mapsto 2\\ 2 \mapsto 2\end{cases}$$

Then $f \circ f = f$. So $f$ acts as the identity morphism and is idempotent, an isomorphism, but not bijective.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.