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Question:

Let $A$ and $B$ be arbitrary sets, with $\alpha:A\rightarrow B$ an injection. Show how to define $\beta:B\rightarrow A$ such that $\alpha \beta$ is the identity function on $A$.

Solution:

For $a\in A$ define $(a\alpha)\beta=a$. For $b\in B$, $b\neq a\alpha$ for any $a$, define $b \beta=a_{1}\in A$.

Source: Groups: A Path To Geometry by R. P. Burn. Chapter: 1 Question: 24

The injection has the property $x\alpha=y\alpha \Rightarrow x=y$.

My problem lies in understanding this statement "For $b\in B$, $b\neq a\alpha$ for any $a$, define $b \beta=a_{1}\in A$".

Does "$b\neq a\alpha$ for any $a$" mean that no image of any $a\in A$ can be equal to itself? Why must this be true?

shredalert
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1 Answers1

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I’ll rephrase the solution in what I hope is a more understandable way.

We need to define $b\beta$ for each $b\in B$. There are two kinds of elements of $B$: those that are in the range of $\alpha$, and those that are not. If $b=a\alpha$ for some $a\in A$, we define $b\beta=a$. Now fix a particular $a_1\in A$. If there is no $a\in A$ such that $b=a\alpha$, we define $b\beta=a_1$.

The last two sentences correspond to the statement about which you’re asking.

Brian M. Scott
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    So putting it into words. As long as there are elements of $A$ which map to $B$ under $\alpha$, those images get mapped back to the original elements $a\in A$ as you have stated $b\beta=a$.

    For all the elements of $B$ which are not images of the elements in $A$ we assign a particular element $a_{1}\in A$ which serves as a kind of 'dummy image'. Doesn't this also mean that $\beta$ is surjective, as for any $a\in A$ there is a $b\in B$ such that $b\beta=a$?

    The solution the book I am studying from provided mislead me in its meaning. Many thanks for such a clear explanation, Brian. cont...

     – shredalert Jan 13 '17 at 20:58
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    ...cont.. Could you please check my comment to see if I understood things properly? – shredalert Jan 13 '17 at 20:58
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    @shredalert: Your understanding is correct, and yes, $\beta$ is surjective. You’re welcome! – Brian M. Scott Jan 13 '17 at 21:00