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this question is really causing me to pull my hair out. I have to find a function $f : A \to B$ such that all of the three conditions are true for the same function $f$:

(1) there is a function $f_1 : B \to A$ so that $f_1 \circ f = \mbox{id}_A$ but $f \circ f_1 \ne \mbox{id}_B$.

(2) there is a function $f_2 : B \to A$ so that $f \circ f_2 = \mbox{id}_B$ but $f_2 \circ f \ne \mbox{id}_A$.

(3) there are different functions $f_3$, $f_4 : B \to A$ so that $f \circ f_3 = \mbox{id}_B$ and $f \circ f_4 = \mbox{id}_B$.

I have tried things like root $x$ and $x^2$ with modulus (absolute value) on difference sides, but, I can't seem to get anything that applies for all $3$ conditions. I would really appreciate some hints or help with this!

Thanks, Helen

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  • $\begingroup$ See, I have to find the actual functions... I'm not sure how to get to that point where it works for all of them. $\endgroup$ Oct 17, 2014 at 11:28
  • $\begingroup$ @Did that would make conditions 1 and 2 incompatible. $\endgroup$
    – Arthur
    Oct 17, 2014 at 11:29

2 Answers 2

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If $f_1\circ f=\operatorname{id}_A$ and $f\circ f_2=\operatorname{id}_B$ then

$$f_1=f_1\circ \operatorname{id}_B=f_1\circ (f\circ f_2)=(f_1\circ f)\circ f_2=\operatorname{id}_A\circ f_2=f_2$$ So the conditions cannot be met.


If there is a right- and a leftinverse then they will coincide. In that case we can speak of an inverse, and inverses are unique.

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  • $\begingroup$ Thanks, I knew there must have been something wrong. I spent ages at it. I must ask the lecturer about it. Thank you for the help $\endgroup$ Oct 17, 2014 at 11:47
  • $\begingroup$ Is your hair still on your head? I sincerely hope so! :) $\endgroup$
    – drhab
    Oct 17, 2014 at 11:49
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It's not possible to have both left and right inverse but they are not equal.

Suppose that $f(f_2(y))=y$ and $f_1(f(x))=x$, then $f_1(y)=f_1(f(f_2(x))=f_2(y)$. So they have to be the same, which means $f_2$ is left inverse, and $f_1$ is right inverse.

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  • $\begingroup$ Sorry, I'm confused. Are you saying that the question must have an error? $\endgroup$ Oct 17, 2014 at 11:36
  • $\begingroup$ @HelenByrne Yes $\endgroup$
    – John
    Oct 17, 2014 at 11:42

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