1. ## Left Inverse

For each of the mappings f given in Excercise 1, determine whether f has a left inverse. Exhibit a left inverse whenever one exists

b. $f(x) = 3x$

I know the function is one to one and has a left inverse as follows

$\begin{array}{rl}
f(m) = f(n) \implies & 3m = 3n\\
\implies & m = n \end{array}$

Based on the lemma we have (lemma 1.24 in our book we know the following)

If there is an element $y \in Z$ such that $f(y)=x$, then $g(x)= y$.
$f(x) = 3x$
$f(y) = 3y$
$x = 3y$
$\frac{x}{3} = y$
$g(x) = \frac{x}{3}$

Now I know that is the inverse of the function but is that the inverse only when x is a multiple of 3? and it's an arbitrary $a_{0}$ for all other conditions? How would I write that ...

A left inverse $g: Z \rightarrow Z$ is $$g(x) = \left\{ \begin{array}{ll} \frac{x}{3} & \mbox{if x is a multiple of 3}\\ 1 & \mbox{if x is not a multiple of 3}\end{array} \right.$$

As you can see the problem is essentially solved it's more or less how do I format my answer that would be expected by a professor when I take this course next semester.

2. Originally Posted by imind
As you can see the problem is essentially solved it's more or less how do I format my answer that would be expected by a professor when I take this course next semester.
I doubt that any of us could possibly know what any particular professor expects of you. You did not even tell the name of the course or the level.

3. It's modern algebra ... 3000 level or junior level ...

Does anyone know if the above I wrote is actually correct? (the course shouldn't be relevant to if the information is right or not ...)

4. Found the answer I needed ... it is indeed

$$g(x) = \left\{ \begin{array}{ll} \frac{x}{3} & \mbox{if x is a multiple of 3}\\ 1 & \mbox{if x is not a multiple of 3}\end{array} \right.$$

1 can be any arbitrary value you choose.