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The following is an exercise from Jacobson's Basic Algebra I:

Let $S=\{1,2,\cdots\}$. Give an example of two maps $\alpha, \beta$ of $S$ into $S$ such that $\alpha \beta=1_{S}$ but $\beta \alpha\ne 1_{S}$. Can this happen if $\alpha$ is bijective?


I have a fairly good answer for this exercise, I think.

An Example

Let $\alpha(s)=|s-2|$ and $\beta(s)=s+2$ for $s \in S$. Thus, $(\alpha \circ \beta)(s)=|(s+2)-2|=|s|=s$, which implies $\alpha \beta=1_{S}$.

However, $(\beta \circ \alpha)(s)=|s-2|+2$. We see that $(\beta \circ \alpha)(1)=3$, hence $\beta \alpha\ne 1_{S}$.

Consequences of $\alpha$ Being Bijective

If $\alpha$ is bijective, there exists an inverse map $\gamma$ such that $\gamma \alpha=\alpha \gamma=1_{S}$. Since $\alpha \beta=1_{S}$, we have that $\alpha \beta=\alpha \gamma$. Composing both sides of the equation with $\gamma$, we have that $\gamma(\alpha \beta)=\gamma(\alpha \gamma)$. Next, using the law of associativity, we have that $(\gamma \alpha)\beta=(\gamma \alpha)\gamma$. Since $\gamma \alpha=1_{S}$, we thus have $1_{S}\beta=1_{S}\gamma$. Hence, $\beta=\gamma$.

Since we previously stated that $\gamma \alpha=1_{S}$ and we've shown that $\gamma=\beta$, it must be so that $\beta\alpha=1_{S}$. $\square$


My question

Are there more natural examples of $\alpha$ and $\beta$? Furthermore, is my proof accurate? I ask the second question only as I see it directly pertains to the first.

P.S. By natural, I mean elegant or insightful.

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    There's nothing wrong with coming with $a$ solution that is not "aesthetically optimal" at first. The aesthetic solutions usually come later after you've had more time to fully internalize everything a$b$out a pro$b$lem. But there c$a$n be as much learning as finding a non-standard solution as there is to finding a very elegant one. BTW, I would consider the example I presented as "geometric" in so far as it can be visualized as pulling a string of beads in a one-dimensional universe. :)2012-10-21

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$\alpha(x)=2x$ and $\beta(x) = \lfloor \frac{x+1}{2} \rfloor$ is probably more natural.

More generarily, given any $\alpha$ which is 1-1 but not bijective function, you can always find a $\beta$ so that $\beta \alpha =1_S$. But since $\alpha$ is not bijective, $\alpha \beta \neq 1_S$.

P.S. Your proof looks OK to me.

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    @Henry, missed that, ty. Fixed.2012-10-22