Proving why $\Bbb Z_{4} \rightarrow \Bbb Z_{6} \text{ given by } f(\overline x) = [2x+1] $ is not a function.

Question

Question presented: Is following a function from the indicated domain to the indicated co domain?

$f:\Bbb Z_{4} \rightarrow \Bbb Z_{6} \text{ given by }$ $ \bbox[white,1px,border:1px solid red]{f(\overline x) = {\left [ 2x+1\right]} } $

Q.S: So my main question for this problem is how does the co domain $f(\overline x) = [2x+1] $ affect the equivlanece class $\Bbb Z_{4}$?

QS: So according to my work here would it be correct to assume that $\Bbb Z_{4}$ would change because of the codomain.

Q.S: Lastly how does $x =y(mod)m$ affect the final result of the problem?

In this case we represent an element of the domain as an $\bar x$ and use the notation $[x]$ for equivalence classes in the co-domain.

The set of equivalence classes for the relation $\cong_{m}$ is denoted $\Bbb Z_{m}.$

$f(\overline0) = [1],f(\overline1) = [3],f(\overline2) = [5],f(\overline3) = [7],f(\overline4) = [9],f(\overline5) = [11],f(\overline6) = [13] $

$(4x+0)*2+1 \;\text{ This is what I am doing to get the result for the work bellow for } \Bbb Z_{4} $ $x$ is an integer that is being multiplied to get the result it is a visual representation for the function.

$\Bbb Z_{4},\; \overline 0 = \{ ...,-15,-7,1,9,17,... \} \; \Leftarrow \Rightarrow(4x+0)*2+1 $

$\overline 1 = \{ ...,-13,-5,3,11,19.., \} \Leftarrow \Rightarrow \; (4x+1)*2+1$

$\overline 2 = \{ ...,-11,-3,5,13,21,... \}$

$\overline 3 = \{ ...,-9,-1,7,15,23..., \}$

$\Bbb Z_{6}, \; \overline 0 = \{ ...,-12,-6,0,6,12,... \} \Leftarrow \Rightarrow \qquad (6x+0) $

$\overline 1 = \{ ...,-11,-5,1,7,13,... \} \Leftarrow \Rightarrow \qquad (6x+1)$

$ \overline 2 = \{ ...,-10,-4,2,8,14,... \} \Leftarrow \Rightarrow \qquad (6x+2)$

$ \overline 3 = \{ ...,-9,-3,3,9,15,... \} $

$ \overline 4 = \{ ...,-8,-2,4,10,16,... \}$

$ \overline 5 = \{ ...,-7,-1,5,11,17,... \}$

The reason for this not being a function is because $\Bbb Z_{6} \; \overline 0 \neq \overline4 \; $ or $[1] \neq [9] $

Answer 1

I'm going to answer the question in the title:

The function $f: \mathbb{Z} \to \mathbb{Z}/6$, $x \mapsto [2x+1]$ does not descend to a function on $\mathbb{Z}/4$ because $x_1 \cong x_2 (mod \,\, 4) \nRightarrow f(x_1) \cong f(x_2) (mod \,\, 6)$. You can check this explicitly for the choices $x_1=0$, $x_2=4$.

Answer 2

It is an onto relation from a set of lesser to a set of greater cardinality. Thus it is not a function by definition of cardinality.

Answer 3

$1\equiv5\pmod4$, yet $f(1)=3\not\equiv f(5)=11\pmod6$

That is, $f(x)=2x+1$ is not well defined as a function from $\mathbb{Z}_4$ to $\mathbb{Z}_6$; choosing a different representative for $x\bmod4$ gives a different residue class for $f(x)\bmod6$.