0

Let $A$ be a set of at least $2$ elements and $M=\left\{q\mid q:A\rightarrow A\text{ is a function}\right\}$.

$($a$)$ $f\in M$ is left absorbing if and only if $f$ is a constant function.

$($b$)$ $M$ has no right absorbing elements.

For part $($a$)$, I know that if $f$ is left absorbing, then $f*g=f$ for all $g\in M$. Here, $*$ is an arbitrary binary operation. Then, I have that $f*f=f$. But, I am not sure where to go from here. I am a bit confused since I don't know what $f*f=f$ really says about $f$ (other than that it is idempotent) and since $*$ is not specified. I feel as if I can get part $($b$)$ once I understand part $($a$)$ better.

  • 2
    Are you sure $*$ isn't just function composition $\circ$? Consider what $f\circ g=f$ tells you when $g$ is a constant function. – Dap Sep 18 '17 at 19:09

1 Answers1

1

For a) showing that if $f$ is constant then $f$ is left absorbing is trivial. To show that if $f$ is left-absorbing, then $f$ is constant we will show, that if $f$ is not constant, then it's not left-absorbing.

So assume, that $f$ is not constant, ie. there are two elements $a,b$ in $A$, such that $$f(a)\neq f(b)$$ Now let's take $g\in M$, such that $g(a)=b$. We have then: $$f(g(a))=f(b)\neq f(a)$$ Thus $f$ is not left-consuming.

To show b) let's assume, that $M$ contains function $f$ that is right-absorbing, ie. $\forall g\in M\forall x\in A g(f(x))=f(x)$.

Let's say, that $f(a)=b$.

Let's take a function $g\in M$ such that $g(b)\neq b$. We have then: $$g(f(a))=g(b)\neq b = f(a)$$ Thus $f$ is not right-consuming.

Jaroslaw Matlak
  • 4,895
  • Thank you for the answer. I guess the key here was knowing that the binary operation was just function composition. Thanks again! –  Sep 18 '17 at 19:24
  • What a coincidence! I was just going to suggest a simplification, but your edit just did it. Cheers! –  Sep 19 '17 at 04:37