Homomorphism or not? Consider the function
$$\phi :\begin{align} \mathbb Z_4 &\to \mathbb Z_4\\z &\mapsto 1\end{align}$$ 
Why is this not a group homomorphism?
On the other hand
Why $$\psi :\begin{align} \mathbb Z_4 &\to \mathbb Z_4\\z &\mapsto 2z\end{align}$$ 
for all $z\in Z_4$ is a group homomorphism?
My guess was that because only the identity can be linked to the identity, but in our case 4 elements are linked to the identity.
For case 2, simply check all possible combinations to see if $f(z_1z_2)=f(z_1)f(z_2)$
 A: call the function $f:Z_4\rightarrow Z_4$, $f(a)=1$.
For something to be a homomorphism we must have $f(a+b)=f(a)+f(b)$. So just for a counterexample choose 1 and 2. we have
$f(1+2) = f(3) = 1$
but,
$f(1)+f(2)=1+1=2$
so this isn't a homomorphism. Now check the same for mapping $f(a)=2a$.
$f(a+b)=2(a+b)=2a+2b=f(a)+f(b)$
therefore that is a homomorphism.
A: I thinkg you're considering $\mathbb{Z_4} $ as additive group. It is known that if $\varphi: G \to H$ is a group homomorphism, then $$\varphi(1_G) = 1_H$$
Just write $1_G = 1_G1_G$, then $\varphi(1_G) = \varphi(1_G)\varphi(1_G)$. Follows that $\varphi(1_G) = 1_H$.
In this case, $G = H = \mathbb{Z}_4$ and therefore $1_G = 1_H = 0$. Since the function $x \mapsto 1$ does not map $0$ to $0$, it could not be a group homomorphism.
For case two, you should take two arbitrary elements and check that they satisfy the condition of group homomorphism. 
A: Hint:


*

*Case one, notice that $\varphi (\overline{2}) = \overline{1}$ is not possible because the order of $\overline{1}$, which is $4$, does not divide $2$,  the order of $\overline{2}$. 

*Case two, the fact that $\mathbb Z_4$ is abelian settle things down.
A: You have to be careful, when talking about groups, to make sure it is clear what operation is being used. In this case, it appears from context that the group $(\Bbb Z_4,+)$, is intended.
Your intuition about the image of the identity is correct: in fact, we have the following:
For ANY two groups $(G,\ast)$ and $(H,\ast')$ and any group homomorphism $f:G \to H$, we have: $f(e_G) = e_H$. To see this, note that for any $h \in f(G)$ (so that $h = f(g)$ for some $g \in G$):
$f(e_G)\ast' h = f(e_G)\ast' f(g) = f(e_G \ast g) = f(g) = h$,
$h \ast' f(e_G) = f(g) \ast' f(e_G) = f(g \ast e_G) = f(g) = h$.
This tells us that $f(e_G)$ acts as the identity of $f(G)$, and since $f(G)$ is a subgroup of $H$, they have the same identity, namely $f(e_G) = e_H$.
So for your first example, since $f([0]) = [1] \neq [0]$, $f$ cannot be a homomorphism.
Now, in your second example, we DO have $f([0]) = 2[0] = [2]\cdot [0] = [2\cdot 0] = [0]$, but this alone is not enough to prove we have a homomorphism. We need to verify:
$f([k] + [m]) = f([k]) + f([m])$.
So, starting with the LHS:
$f([k] + [m]) = f([k + m]) = 2[k+m] = [2]\cdot[k + m] = [2(k+m)]$
Now, inside the brackets, we have "ordinary integers", and can use the rules we know about such:
$[2(k+m)] = [2k + 2m] = [2k] + [2m] = [2]\cdot[k] + [2]\cdot[m]\\ = 2[k] + 2[m] = f([k]) + f([m]).$
