Category Theory - Isomorphisms of Posets

[Math Amateur]

I am beginning to read "Category Theory: Second Edition" by Steve Awodey.

On page 12 (see attachment) he defines isomorphisms as follows:

-------------------------------------------------------------------------------

Definition 1.3. In any category,$$ C $$, an arrow $$f \ : \ A \to B $$ is called an isomorphism, if there is an arrow $$ g \ : \ B \to A $$ such that

$$ g \circ f = 1_A \text{ and } f \circ g = 1_B $$

Since inverses are unique (proof?), we write $$ g = f^{-1} $$. We say that A is isomorphic to B, written $$ A \cong B $$, if there exists an isomorphism between them.

----------------------------------------------------------------------------

Awodey then writes:

"... ... (this definition) has the advantage over other possible definitions that it applies in any category. For example, one sometimes defines an isomorphism of sets (monoids etc.) as a bijective function (respectively homomorphism) , that is, one that is "1-1" and onto - making use of the elements of the objects. This is equivalent to our definition in some cases, such as sets and monoids. But note that in the category of posets, Pos, the category theoretic definition gives the right notion, while there are "bijective homomorphisms" between non-isomorphic posets. ... ..."

BUT ... I thought previously that an isomorphism was defined as a bijective homomorphism ... but it seems this is only suitable for some contexts ... ?

I cannot imagine two posets with a bijective homomorphism between them that is not an isomorphism ... can someone come up with an example of such a case ...

Further, since I do not know much about posets can someone supply a good reference - online notes or text? (Awodey is threatening to use posets and monoids as examples throughout the text!)

Peter

 

[ModusPonens]

I'm sorry if this is absurd, but I didn't read with complete attention, so bear with me. The problem is, I think, that although bijective homomorphisms are isomorphisms (in the categorical sense, isomorphisms (in the categorical sense) are not necessarily bijective homomorphisms.

An example of an morphism that is epi, but not surjective, is a homomorphism from $\mathbb{Z}$ to $\mathbb{Q}$, in the category of rings with morphisms that preserve the multiplicative identity. So it's not that strange that there are morphisms that are iso, but are not bijective functions.

 

[ThePerfectHacker]

In category theory the collection of morphisms between two objects is denoted by $\text{hom}(A,B)$. This can possibly be empty. We typically think of $\text{hom}(A,B)$ as the homomorphisms from $A$ to $B$. But this is abstract. It is just a collection of stuff, called morphisms, associated for every pair $(A,B)$.

 

[Math Amateur]

I'm sorry if this is absurd, but I didn't read with complete attention, so bear with me. The problem is, I think, that although bijective homomorphisms are isomorphisms (in the categorical sense, isomorphisms (in the categorical sense) are not necessarily bijective homomorphisms.

An example of an morphism that is epi, but not surjective, is a homomorphism from $\mathbb{Z}$ to $\mathbb{Q}$, in the category of rings with morphisms that preserve the multiplicative identity. So it's not that strange that there are morphisms that are iso, but are not bijective functions.

Thanks ModusPonens ... but given what you say I am having trouble processing Awodey's remark as follows:

" ... ... there are bijective homomorphisms between non-isomorphic posets ..."

Doesn't this mean that for posets, a bijective homomorphism is not (necessarily) an isomorphism?

Or am I making some logical error?

Peter

 

[ThePerfectHacker]

" ... ... there are bijective homomorphisms between non-isomorphic ..."

Here is an example from topology.

Consider category Top, it consists of all topological spaces and the morphisms between two topological spaces are continuous maps between them. Two topological spaces are called homeomorphic if they are isomorphic in the categorical sense. It is not true however that if $f:X\to Y$ is a continuous bijection between topological spaces that necessarily $f^{-1}:Y\to X$ is continuous as well.

 

[ModusPonens]

Thanks ModusPonens ... but given what you say I am having trouble processing Awodey's remark as follows:

" ... ... there are bijective homomorphisms between non-isomorphic posets ..."

Doesn't this mean that for posets, a bijective homomorphism is not (necessarily) an isomorphism?

Or am I making some logical error?

Peter

You're correct. And Hacker's example is perfect. Sorry for complicating this.

 

[Deveno]

It is important to realize that "arrows" are not always "functions".

For example, in a poset with the unique arrow $f:a \to b$ defined iff $a \leq b$, the arrows are not functions of any sort.

Another interesting example is the following category, where the collection of objects is the set of positive integers, and an arrow:

$f:k \to m$ is a $m\times k$ matrix with entries in a given commutative ring.

*********

A homomorphism between two posets is defined to be an order-preserving map. So consider the map $\phi:\Bbb N^+ \to \Bbb N^+$ defined by:

$\phi(k) = k$, which is clearly bijective.

Order the first copy by divisibility and the second copy by the usual "less than or equal to".

It follows that $\phi$ is an order-preserving map, since if:

$a|b$ we have $a \leq b$.

However, $\phi^{-1}$ is NOT an order-preserving map, since $a \leq b$ does not imply $a|b$ (take $a = 2, b = 3$), so it is not a "homomorphism of posets".