Inverse elements for monoid of all injective real functions together with defined op

[missandtroop]

Hello,
learning for my algebra exam I came across a task I cannot wrap my head around for some time already.

The task is to decide weather formula (not sure with English terminology here, sorry) (f * g)(r) = f(g(r) - 1), for f,g: R -> R and r \in R, defines operation on set S of all injective real function, such that (S,*) is semigroup, group respectively.

So far I what I've done is that I had shown that * is a function from S^2 -> S like so:



Next, I've shown, using the function h from above that * is associative:

which should make (S, *) a semigroup.

Then I found a neutral element e(r)=r+1, r \in R :

existence of which should imply (S,*) is a monoid.

Lastly, to decide whether it's group I need to decide whether every element in S has an inverse (meaning that for each f \in S there is som g \in S that f*g = e and g*f = e) or not. This is where/when my brain stopped working (partially because it's 3 AM here but I cannot sleep without finishing this one). I cannot find way to show that there is such g for any f, nor the opposite.

Any hints, please? :smile:
Thank you.

P.S. this is my first post here so I hope I formatted the question OK. Have a nice day all of you.

 

[barrick]

Hi missandtroop,

This seems correct so far (although this is obvious, you should state that e is injective, i.e., that it belongs to the set S).
Note that h is bijective and e is the inverse of h.

If you allow non-surjective functions, you will not have a group.

To see this, let us take \(\displaystyle f(x)=e^x\); this function is injective but not surjective. If g is a right inverse for f with respect to *, we must have:

\(\displaystyle
\begin{align*}
(f*g)(x) &= f(g(x)-1)\\
&= e^{g(x)-1}\\
&= x+1
\end{align*}
\)​

for all x. As \(\displaystyle e^{g(x)-1}>0\) no matter how you define g(x), this is impossible if x < -1.