4
$\begingroup$

Definition Let $A$ and $B$ be any objects in category $C$. We say that $A$ is retract of $B$, notated by $A\le_r B $, iff $\exists s:A\rightarrow B, r:B\rightarrow A$ such that $r\circ s=1_A$.

It is easy to see that $\le_r$ is reflexive and transitive. But, I am having difficulty seeing that $\le_r$ is also anti-symmetric. In other words, is following statement true:

$$[(A\le_r B)\wedge (B\le_r A)]\Rightarrow (A\cong B).$$

Motivation behind this question stems from concept of "least co-separator". An example of least co-separator is 2-set in category of sets and functions. One may define a least co-separator as a co-separator A such that $\forall co-separator B\in C, A\le_r B$. From this, given that $\le_r$ is anti-symmetric, uniqueness of co-separator (upto isomorphism) follows. I would much appreciate any input on this too.

If the statement about anti-symmetry is false, could you please provide any elementary example, if possible?

$\endgroup$
2
  • $\begingroup$ Antisymmetric or asymmetrical? $\endgroup$
    – manooooh
    May 27, 2018 at 20:31
  • $\begingroup$ @manooooh Sorry, I meant Anti-symmetric. I will correct my question. Thank you! $\endgroup$
    – Jaspreet
    May 28, 2018 at 5:05

1 Answer 1

5
$\begingroup$

It is not antisymmetric in general. For instance, in topological spaces, $A=\mathbb{N}\times \mathbb{R}$ and $B=A\coprod [0,1]$ are each retracts of each other but not homeomorphic. The retraction $A\leq_r B$ is obvious. For the retraction $B\leq_r A$, map $[0,1]$ into one of the copies of $\mathbb{R}$ (in which it is a retract) and pair up the rest of the copies of $\mathbb{R}$ in $A$ with those in $B$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .