I would tend to disagree with the earlier commenter: I think you mean to think of as , where you think of as a function . The reason is that when you’re looking for inverse images for elements of , you’ll find them for $i$ but not necessarily for .

For example, if and is multiplication by 2, then . Now, consider : It has no preimage under , since for all . However, does have (two) preimages under ; namely, the elements and .

Anyway, to be precise: We want to define , so given , we have to define . Since , choose such that . Then we define .

To make this work, we have to show two things: first, that is well-defined, and second, that for all so that for all .

Note: I do think that where you put , you probably meant .

Thanks for this helpful post!

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