Xrfgtjtk

This is probably an elementary question, but I'm new to this machinery.

Let $G$ be a group and $N$ be a normal subgroup of $G$. Let $Gamma$ be another group. Suppose we have a homomorphism $phi:Gamma to G/N$.
Now we have the homomorphism $phi:Gamma to G/N$ and the canonical quotient homomorphism $psi:G to G/N$.

Denote by $H$ the pullback of the two homomorphisms to $G/N$. That is, $H subseteq G times Gamma$ comprising elements $(g,gamma)$ such that $phi(gamma)=psi(g) in G/N$.

So we have the following two diagrams:
$$1 rightarrow N rightarrow Gstackrel{psi}{rightarrow} G/Nrightarrow 1$$
which is a short exact sequence, and another diagram
$$begin{array}
GG & stackrel{psi}{longrightarrow} & G/N\
uparrow & & uparrow{phi} \
H & longrightarrow & Gamma
end{array}
$$
Is the following also a commutative diagram?
$$begin{array}
11 & longrightarrow & N & longrightarrow & G & stackrel{psi}{longrightarrow} & G/N & longrightarrow & 1\
& & uparrow & & uparrow & & uparrow{phi} & & \
1 & longrightarrow & N & longrightarrow & H & longrightarrow & Gamma & longrightarrow & 1\
end{array}
$$
That is, can $N$ be embedded in the pullback $H$?

Secondly, suppose the map $phi:Gamma to G/N$ can be lifted to a map $phi':Gamma to G$ so that we have a commuting diagram bypassing $H$. In this case, is $H$ a semidirect product of $N$ and $Gamma$?

In other words, is the bottom exact sequence split when $phi$ can be lifted diagonally to $G$, and is the converse true too?

Thanks!

Yes and Yes. The subgroup $N$ of $H$ that you are looking for is ${(n,1) : n in N }$, and the complement you are looking for is ${(phi'(gamma),gamma) : gamma in Gamma}$.