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I know how to determine injectivity and surjectivity for maps between regular sets, but in this case I've got some problems. How can I solve this?

Given the following map $\psi:\overline{x} \in \mathbb{Z}_{16}\mapsto \overline{7}\overline{x}\in\mathbb{Z}_{16}$. Without calculating a single element's image, and just using the properties of $\overline{7}$ in $\mathbb{Z}_{16}$, decide if $\psi$ is injective, surjective or both. If possible, find the inverse of $\psi$.

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$7\times 7=49=3\cdot 16 +1$ so what about $\psi\circ\psi$? –  Davide Giraudo Feb 13 '12 at 12:43
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Note also that injectivity and surjectivity are equivalent for functions between finite sets of the same cardinality. –  lhf Feb 13 '12 at 12:58
    
I haven't tried nothing, cause I don't know how to proceed. I was able to do inj. and surj. check calculating every single elements. But in this case I can't do that. @davide: what do you mean? Why $7\times 7$? –  Mariano Feb 13 '12 at 13:04
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We have $\bar 7 \cdot\bar 7=\bar 1$. Thanks to that you can say what $\psi\circ\psi$ is. –  Davide Giraudo Feb 13 '12 at 13:12
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@J.D., the bar means class mod 16. –  lhf Feb 13 '12 at 19:41

2 Answers 2

You may want to attack the more general question of what you can say about $\psi_a:x\mapsto ax$. Clearly, $\psi_1 = id$ and $\psi_a \circ \psi_b = \psi_{ab}$. In particular, if $ab \equiv 1$ then $\psi_a$ and $\psi_b$ are inverses of each other.

Now, consider whether there is a $b$ such that $7b\equiv 1 \bmod 16$.

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Since $7$ is relatively prime to $16$, it is a unit in the ring $\mathbb{Z}_{16}$, so $7x=7y$ implies $x=y$. Thus, multiplication by $7$ is injective. Since $\mathbb{Z}_{16}$ is a finite set, multiplication by $7$ is also bijective. The inverse of the map is also multiplication by $7$ since $7$ is its own inverse mod $16$.

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