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

  1. #1

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

    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:

    Screen Shot 2018-01-15 at 02.52.34.jpg

    Next, I've shown, using the function h from above that * is associative:
    Screen Shot 2018-01-15 at 02.58.31.png
    which should make (S, *) a semigroup.

    Then I found a neutral element e(r)=r+1, r \in R :
    Screen Shot 2018-01-15 at 03.01.44.png
    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?
    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.

  2. #2
    New Member
    Join Date
    Nov 2017
    Brussels, Belgium
    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 [tex]f(x)=e^x[/tex]; this function is injective but not surjective. If g is a right inverse for f with respect to *, we must have:

    (f*g)(x) &= f(g(x)-1)\\
    &= e^{g(x)-1}\\
    &= x+1

    for all x. As [tex]e^{g(x)-1}>0[/tex] no matter how you define g(x), this is impossible if x < -1.
    Last edited by barrick; 01-15-2018 at 07:27 AM.

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