# Non-bijective isomorphism in a category of of sets.

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 Set.

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

Proof

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.

• 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? Sep 5 '14 at 18:42
• 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)). 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.
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.