Map factor through quotient map? Affine coordinate axes in $\mathbb{A}^2(\mathbb{C})$ (place where $x=0$ or $y=0$), so $\simeq \mathbb{C}[x,y]/\langle xy\rangle=:A$.
$\forall a,n\in\mathbb{N}-\{0\}$: Do all ring homomorphism $f:A\to \mathbb{C}[x]/\langle x^a\rangle$ have ring homorphism $f':A\to \mathbb{C}[x]/\langle x^{a+n}\rangle$ such that $f=\pi \circ f'$ where $\pi:\mathbb{C}[x]/\langle x^{a+n}\rangle\to \mathbb{C}[x]/\langle x^{a}\rangle$ is quotient map? What if we change $\mathbb{C}$ for different field?
Regards,
Bai
 A: The answer is no for every  field   $k$. Here is a counterexample:   
Define  $A=k[X,Y]/\langle XY\rangle=k[x,y],\; k[T]/\langle T^2\rangle=k[\epsilon ]$ and $k[Z]/\langle Z^3\rangle=k[\zeta ]$, so that $xy=0, \epsilon ^2=0, \zeta^3=0$
Define  the $k$-algebra morphism $f: k[x,y]\to k[\epsilon ]$ by demanding $f(x)=f(y)=\epsilon$ .
 I claim that this morphism $f$ cannot be lifted to $f':A\to k[\zeta ]$. 
Indeed such a lift would have to satisfy $f'(x)=\zeta+a\zeta^2, f'(y)=\zeta+a'\zeta^2$ (for some $a,a'\in k$), since the only elements of $k[\zeta ]$ sent to $\epsilon$ by the quotient map $\pi$ are of the form $\zeta+a\zeta^2$.
But this leads to  $0= f'(0)=f'(xy)= f'(x)f'(y)=(\zeta+a\zeta^2)(\zeta+a'\zeta^2)=\zeta^2\neq 0$, a contradiction.  
Edit
As an answer to the OP's request, here are some supplementary details.
Let me remind you that the $k$-algebra  morphism   $\pi: k[\zeta]\to k[\epsilon]$ is given by $\pi(c+b\zeta+a\zeta^2)= c+b\epsilon$, since it is induced by $\Pi:k[Z]\to k[T]:Z\mapsto T$.
The key point is that  $\pi(\zeta^2)=\epsilon ^2=0$ .
The condition $\epsilon=\pi(c+b\zeta+a\zeta^2)=c+b\epsilon \in k[\epsilon]$  then forces $c=0$ and $b=1$, so that indeed any element of  $k[\zeta]$ sent by $\pi$ to $\epsilon$ must be of the form $\zeta+a\zeta^2$ .
