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Let $M = \mathbb{N} \backslash \{1\}$ and $f: \mathbb{N}\rightarrow M$ be a mapping with $f(n) = 2n$

Let $g: M \rightarrow \mathbb{N}$ be the mapping with

$$g(n) = \begin{cases} \dfrac{n}{2} & \text{if $n$ is even} \\ 2018 & \text{if $n$ is odd} \end{cases}$$

Show that $f$ has a left inverse and it does not have a right inverse and prove that $f$ has infinitely many left inverses.

I know it has a left inverse because $g \circ f = g(2n) = \frac{2n}{2} = n$ so $g$ is an identity map on $\mathbb{N}$. However for the right inverse, $f \circ g = f(\frac{n}{2}) = \frac{2n}{2}$ if $n$ is even but if $n$ is odd, then $f(2018) = 4036$

Is this enough to show that there isn't a right inverse?

Thanks for the help!

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    $\begingroup$ correct your misprints like : g is an inverse ; and your composition of g and f is evaluated at n $\endgroup$ – StuartMN Sep 4 '18 at 3:14
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    $\begingroup$ You showed g is not a right inverse ,not that there is no right inverse .Try to show that if there is a left and a right inverse then they are the same . $\endgroup$ – StuartMN Sep 4 '18 at 3:18
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    $\begingroup$ Change $g$ to $g(n)=2017$ if $n$ is odd. Is this also a left inverse? Can you think of infinitely many now? $\endgroup$ – amsmath Sep 4 '18 at 3:21
  • $\begingroup$ "As for proving that there are infinitely many left inverses, it would just be the set of even numbers, which is infinite."—this shows a misunderstanding: a left inverse is a function, not a set. You are being asked to show that $g$ isn't the only left inverse but, rather, that there are infinitely many functions (including $g$) each of which is a left inverse of $f$. $\endgroup$ – Greg Martin Sep 4 '18 at 3:22
  • $\begingroup$ @amsmath thank you, so basically, $f$ has infinitely many left inverses and they can all be of the same type as $g$ except with an infinite amount of numbers if $n$ is odd, right? but how can I start to prove it $\endgroup$ – user8290579 Sep 4 '18 at 3:49
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Suppose there were a right inverse $h$, that is, some $h:M\longrightarrow \Bbb N$ such that $f\circ h:M\longrightarrow M$ is the identity. Let $g$ be any left inverse, so that that $g\circ f:\Bbb N \longrightarrow \Bbb N$ is the identity. Then:

$$g = g\circ\big(f\circ h\big) = \big(g\circ f\big)\circ h = h.$$

Notice that this argument does not depend particularly on this $f$, $M$ or $\Bbb N$.

Corollary: If $f:A\longrightarrow B$ admits both a left inverse and a right inverse, then they are equal and unique.

This is what we call the inverse of $f$, and denote it by $f^{-1}$.


Since you've shown that $g$ (which is a left inverse) is not a right inverse, it follows that $f$ has no right inverse.

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  • $\begingroup$ Thanks, I understand that we started off with the supposition that $h$ is a right inverse but I don't understand how showing that $g$ is left inverse and not a right one, implies that there is no right inverse $\endgroup$ – user8290579 Sep 4 '18 at 3:58
  • $\begingroup$ You've shown that $g$ is a left inverse. The argument above shows that if there were also a right inverse $h$, then $g=h$. But because you've shown that $g$ is not a right inverse, we must conclude that there is no right inverse. $\endgroup$ – Fimpellizieri Sep 4 '18 at 4:10
  • $\begingroup$ oh! I see, thank you so much! $\endgroup$ – user8290579 Sep 4 '18 at 4:11
  • $\begingroup$ do you know how I can prove there are infinitely many left inverses? Someone above explained why but I'm stuck on how to start the proof $\endgroup$ – user8290579 Sep 4 '18 at 4:12
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    $\begingroup$ Yes. For each $k\in M$, if you define $$g_k(n) = \begin{cases} \dfrac{n}{2} & \text{if $n$ is even} \\ k & \text{if $n$ is odd} \end{cases}$$ then $g_k$ is a left inverse. Since $M$ is infinite, so are the $g_k$. $\endgroup$ – Fimpellizieri Sep 4 '18 at 4:28

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