Let $f : \mathbb Z\to\mathbb Z$ be $f(n) = 3n + 2.$ Find $g : \mathbb Z \to \mathbb Z$ with $g \circ f$ is the identity function on $\mathbb Z.$

Question

I'm aware that f isn't bijective so it cannot have an inverse function. So g cannot be an inverse function to f. Knowing this, I know some people express such undefined inverses in terms of a piecewise function but I'm not too sure.

Can someone help me out on this please?

I guess if g : Z → Z is defined by $(n-2)/3$ that would be make g ◦ f the identity function on Z, I'm not too sure though.

Edit: I understand that $(n-2)/3$ can't be the inverse to f but how can I work out the title?

$g(n)=\frac{n-2}3$ is not a function $\mathbb Z\to\mathbb Z.$ For example, $g(3)=\frac13.$ — Thomas Andrews, Jan 12 '23 at 21:13
Yes, there is no inverse, but there is a left-inverse. @Vishu (Indeed, there are uncountably many left inverses.) — Thomas Andrews, Jan 12 '23 at 21:15
@ThomasAndrews Yes your right, so how would you construct the g? — Tingo Hugo, Jan 12 '23 at 21:15
The trouble with the obvious inverse is that $n\mapsto (n-2)/3$ does not map integers to integers (e.g., it maps 3 to 1/3). So it only works as an inverse on the image of $f$, rather than on all of $\mathbb Z$. — Semiclassical, Jan 12 '23 at 21:15
@Semiclassical but I don't understand how else you could approach this? — Tingo Hugo, Jan 12 '23 at 21:20

score 3 · Accepted Answer · answered Jan 12 '23 at 21:25

3

Let $K = f(\Bbb Z)$. It's clear that we must have

$$ g(k) = \frac{k-2}{3} \in \Bbb Z $$

for all $k\in K$. What should $g$ be elsewhere? Actually, it doesn't matter, because $g\circ f$ will only feed elements of $K$ to $g$, and so we will get the identity function regardless of what $g$ does to elements of $\Bbb Z - K$.

In other words, for any $h: \Bbb Z-K \to \Bbb Z$,

$$ g(n) = \begin{cases} \frac{n-2}{3} & n\in K \\ h(n) & \text{otherwise} \end{cases} $$

is a solution.

answered Jan 12 '23 at 21:25

Alex Jones

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Thanks. I have a question, I was taught that a function that isn't invertible can have an inverse relation if it's applied to a subset of f's domain. Would g be an inverse relation in this case? – Tingo Hugo Jan 12 '23 at 21:29
There are a lot of ways to turn non-invertible functions into invertible ones by changing the domain or codomain. The domain of $f$ is relevant when it fails to be one-to-one, as we can then restrict to a subdomain which removes the offending duplicates. In this case, however, $f$ failed to be onto, and so we essentially restricted the codomain instead. – Alex Jones Jan 12 '23 at 21:40

Let $f : \mathbb Z\to\mathbb Z$ be $f(n) = 3n + 2.$ Find $g : \mathbb Z \to \mathbb Z$ with $g \circ f$ is the identity function on $\mathbb Z.$

1 Answers1