kernel of quotient is quotient of kernel? Let $A,B,C$ groups and $f:A\to B$ a homomorphism with $C\subset \ker f \subset A$.
Then $f$ induces a map on the quotient $A/C\to B$
Is it then true that $\ker (A/C\to B)= (\ker A\to B)/C~$?
I tried to prove it, but i am not sure if I did it right. The proof seems to tautological to me. Maybe I missed something. If so, what would be a sufficient condition to make the above hold.
Also, does anything change if $A,B$ are topological spaces with $A\subset C$?

So far I came up with this:
If $aC\subset \ker (A/C\to B)$, then $a\in \ker(A\to B)$ and thus $aC \in \ker(A\to B)/C$.
On the opposite: If $aC\in \ker(A\to B)/C$, then $a\in \ker(A\to B)$. But then the image of $aC$ under $A/C\to B$ is $f(a)\in \ker(A\to B)$.

Apart from $f(a)=1_B$ rather than $f(a)\in{\rm Ker}(A\to B)$, your proof looks correct.
When proving an equality like this, you do sometimes find that the proof looks tautological. This is because everything is just true by definition. 
That is ${\rm Ker}(A/C\to B)$ and ${\rm Ker}(A\to B)/C$ both consist precisely of those $aC$ for which $a\in {\rm Ker}(A\to B)$, so they're equal. Note that your if and only if proof is more clear, I've just put it in one sentence to make the point. 
