Let $A$,$B$ be abelian groups. Let $f: A\to B$ be group homomorphism.
Let $C\subset A$ and $D\subset B$ be subgroups of $A$ and $B$.
Let $C\subset D$. There is a canonical map $\tilde{f}: A/C \to B/D,x \bmod C \mapsto x \bmod D$.
I want to prove $\#\ker\tilde{f}\le \#\ker f$.
My try: Since $\ker\tilde{f}=\dfrac{\ker f}{C \cap \ker f}\#\ker\tilde{f}\le \#\ker f$.
However, I have some concerns about this apparent consideration. The reason is that, in the following diagram, it cannot generally be said that $h$ is surjective. Reflecting on this, I wonder if there is an error somewhere in my consideration. That's why I am asking this question here. I would appreciate your help.
$\require{AMScd}$ \begin{CD} \ker f@>>>A @>{f}>> B\\ @V{h}VV @VVV@VVV\\ \ker{\tilde{f}}@>>>A/C @>{\tilde{f}}>> B/D \end{CD}
I assume "Let $C\subset D$" means "Let $f(C)\subseteq D$." (Examples can be done with proper inclusion).
As written, the desired inequality need not hold. Let $C_n$ be the cyclic group of order $n$.
Take $A=C_{4}$, $B=C_2$, $f$ the nontrivial map, $C$ the trivial subgroup, $D=B$. Then $\ker f$ has two elements and $\ker\tilde{f}$ has four elements. So here the kernel of $\tilde{f}$ is larger.
Now Take $A=C_4$, $B=C_2$, $f$ the nontrivial map, $C$ the kernel of $f$, $D$ the trivial subgroup. Now the kernel of $\tilde{f}$ has one element and that of $f$ has two, do here $\ker f$ is larger.
Finally take $C$ and $D$ trivial. Then $\ker f=\ker\tilde{f}$.
So they can be equal, or either one can be larger than the other.