If $f:B\to C$ is a homomorphism and $g:A\to B$ is a monomorphism then is $\ker f\cong \ker f\circ g$? Let $A,B,C$ be groups. Let $f:B\to C$ be a group homomorphism and let $g:A\to B$ be an injective group homomorphism. 

Is it true that $\ker f\cong \ker f\circ g$?

My attempt :
Define $\phi:\ker f\circ g\to \ker f$ by $x\mapsto g(x)$ (Well $\phi$ is basically just $g$ restricted to $\ker f\circ g$). This is clearly a homomorphism and injective, as $g$ is injective. 
It is not clear to me if $\phi$ is surjective or not. 


*

*Is the result false in general?

*Is it true if $f$ is surjective?
Thank you.
 A: The fact is generally false as the following counter-example shows.
Let $f \colon \mathbb Z/4\mathbb Z \to \mathbb Z$ be the null homomorphism (the one such that $f(x)=0$ for each $x \in B$) and let $g \colon \mathbb Z/2\mathbb Z \to \mathbb Z/4\mathbb Z$ be $g(x)=2x$ for each $x \in \mathbb Z$.
Both $f$ and $g$ are group homomorphism and $g$ is injective. Nevertheless $\ker f=\mathbb Z/4 \mathbb Z$ while $\ker f \circ g=\mathbb Z/2\mathbb Z$ which are not clearly isomorphic since they have different cardinality. 
Edit: I see that I did not address the second part of the question, let me make amend for that.
Even if $f$ where surjective it would change nothing: in the counter-example above to replace $f$ with the null homomorphism onto $(0)$ (the null group), it is a surjective homomorphism but the argument goes as before.
Of course things change if you require that $g$ is surjective, but again since $g$ is injective, requiring that $g$ is surjective would imply that $g$ is an isomorphism and so it is pretty obvious that $\ker f \cong \ker f \circ g$.
A: In general, we have:
$\ker(f\circ g)=(f\circ g)^{-1}\{0\}=g^{-1}(f^{-1}\{0\})=g^{-1}(\ker(f))$.
Thus $\ker(f\circ g)\cong\ker(f)$ if $g$ is injective and $\ker(f)\subseteq g(A)$.
A: *

*Yes, because $\ker f\circ g=g^{-1}(g(A)\cap\ker f)$, so it is false, for instance, if $B$ is finite and $g(A)\nsupseteq \ker f$.

*Surjectivity of $f$ has nothing to do with it, since you can always take $C=f(B)$, which leaves $ker$-s unaltered.
