Condition for pullback to split 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:
$$1 \rightarrow N \rightarrow G\stackrel{\psi}{\rightarrow} G/N\rightarrow 1$$ 
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 $\{(\phi'(\gamma),\gamma) : \gamma \in \Gamma\}$.
