$ρ:G\to H$ is group homomorphism, $M$ is a normal subgroup of $G$, $η:G\to G/M$ is the quotient map, $N$ is a normal subgroup of $H$.
$M\subseteq ρ^{-1}(N)$ and $μ:G/M\to H/N$ is the induced homomorphism.
Given some (normal?) subgroup $X$ of $M$, how to construct (normal?) subgroup $Y$ of $G$ so that there is exact sequence $1\to X\to Y\to G/M\to H/N$ where $Y\to G/M$ is just $η:G\to G/M$ restricted to $Y$.
Intuitively, $Y$ is formed by choose an element in each coset in $Ker(μ)$ (see elements of $Ker(μ)$ as cosets of $η:G\to G/M$) for each element in $X$, and the "size" of $Y$ will be like "$|X||Ker(μ)|$", and $Y$ is "compressed $|X|$ times" to be become $Ker(μ)$
Rigorously how to construct $Y$? Also the choice seems not to be unique, is there some canonical choice?
Example
The prototypical example is Pin group:
$V$ is a finite dimensional vector space over field $K$, $Q$ is a non-degenerate quadratic form on $V$, $Cl(V,Q)$ is the Clifford algebra, $Γ$ is the Lipschitz group.
Then $G=Γ$, $M=K^{\times}$, $H=K^{\times}$, $N={K^{\times}}^2$, $X={\pm 1}$, $Y=Pin_V (K)$ the Pin group.
$ρ:G\to H$ becomes $ρ:Γ\to K^{\times}$ which is the spinor norm on $Γ$ defined as $x^t x$ where $x^t$ is the transpose antiautomorphism of $x$.
$η:G\to G/M$ becomes $η:Γ\to Γ/K^{\times}\cong O_V (K)$ where $O_V (K)$ is the orthogonal group induced by $Q$.
$μ:G/M\to H/N$ becomes $μ:Γ/K^{\times}\cong O_V (K)\to K^{\times}/{K^{\times}}^2$ which is the spinor norm on $O_V (K)$.
The exact sequence becomes $1\to {\pm 1}\to Pin_V (K)\to Γ/K^{\times}\cong O_V (K)\to K^{\times}/{K^{\times}}^2$.
Indeed for this prototypical example there may be other $Y$ that is not isomorphic to $Y=Pin_V (K)$, and $Pin_V (K)$ is just one of them defined by elements of $Γ$ that have a spinorm of $\pm 1$.