How to Show that this relation is not well defined.

Question

We represent an element of the domain as an equivalence class $\bar x$, and use the notation $ \left[ x \right]$ for equivalence classes in the codomain. Show that this is not well defined.

$ f: \Bbb Z_{4} \rightarrow \Bbb Z_{6} \; \text{ given by, } f(\bar x) = \left[ 2x+1\right]$

My question is how do does one determine/ show that this is not a function? How does one show that this not one to one?

$f(\bar 0) = 2(0) +1 = [1]$

$f(\bar 1) = [3]$

$f(\bar 2) = [5]$

$f(\bar 3) = [7]$

$f(\bar 4) = [9]$

$f(\bar 5) = [11]$

$f(\bar 6) = [13]$

I also guessing here that if a number is a multiple of another then that may play a role in determining it if is one-one.

Answer 1

$$ f(\overline{7}) = [15] = [3].$$ But $$ f(\overline{3}) = [7] = [1].$$ Since $\overline{7} = \overline{3}$, this is not well-defined.

Answer 2

If we want to get technical.

$\overline 0 = \{a \in \mathbb Z|\exists n \in \mathbb Z: 0 - a = n*4 \}=\{...,-8,-4,0,4,8,12,....\}$

$0 \in \overline 0$ so

$f(\overline 0) = [2*0 + 1] = [1] = \{a \in \mathbb Z|\exists n \in \mathbb Z: 1 - a = n*6 \}=\{..., -5,1,7,13,....\}$

But $0 - 4 = 0-1*4$ so $4 \in \overline 0$.

So $f(\overline 0) = [2*4 + 1] = [9] = \{a \in \mathbb Z|\exists n \in \mathbb A: 9 - a = n*6\} = \{a \in \mathbb Z|\exists n \in \mathbb A: 3 - a = n*6\}=[3]=\{....-9, -3,3,9,15...\}$.

But $[3] = \{a \in \mathbb Z|\exists n \in \mathbb A: 3 - a = n*6\}=\{....-9, -3,3,9,15...\}\ne \{a \in \mathbb Z|\exists n \in \mathbb Z: 1 - a = n*6 \}=[1]=\{..., -5,1,7,13,....\}$

So we have define $f$ in such a way that $f(\overline 0)$ can be interpretted to have at least two incompatible values. Thus $f$ is not well-defined.

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If "$\overline 0 = \{a \in \mathbb Z|\exists n \in \mathbb Z: 0 - a = n*4 \}$" is too complicated looking, we can alternatively say $\overline 0$ is the class = {all multiples of 4} and $\overline x $ is the class = {all integers that are $x$ plus or minus a multiple of 4}. Those definitions are exactly the same.

So $[y]$ is {all integers that are y plus or minus a multiple of 6}

So $f(\overline x) = [2x + 1]$ means "take the set of numbers that are $x$ plus or minus a multiple of 4-- and return that set of all numbers that are $2x + 1$ plus or minus a multiple of 6".

That statement is "not well-defined" because if we take different numbers that are $x$ plus or minus a multiple of 4 (example if we took $x-4$ instead of $x$) we'd get different sets.

Example: if we took $1$ we'd get the set "all numbers that are $3$ ($3= 2*1 + 1$) plus or minus a multiple of $6$" but if we took $5$ (which is $1$ plus or minus a multiple of $4$) we'd get the set "all numbers that are $11$ ($11 = 2*5 + 1$) plus or minus a multiple of $6$". That's a different set.