I am trying to solve the following exercise:
Let
$0 \longrightarrow B \longrightarrow Y \overset{\rho}{\longrightarrow} X \longrightarrow 1 $
and
$0 \longrightarrow A \longrightarrow X \overset{\pi}{\longrightarrow} G \longrightarrow 1 $
be central extensions. Show that the composition
$0 \longrightarrow \mathrm{ker}(\pi \circ \rho) \longrightarrow Y \overset{\pi \circ \rho}{\longrightarrow} G \longrightarrow 1 $
is a central extension of $G$. If $X$ is a universal central extension of $G$, conclude that every central extension splits.
I am lost on the first part of the question. I thought about using the property that for the surjective group homomorphism $\rho$, the center of $Y$ will be mapped into the center of $X$, however I did not make any progress with this idea. For the second claim I understand how I can construct a candidate for a right split using the "universal property" of universal central extensions by applying the result before, but this only delivers a homomorphism $f \colon X \longrightarrow Y$ such that $ \pi = \pi \circ \rho \circ f$ (with the notation from before) and not $\rho \circ f = \mathrm{id}_X$. Any tips are greatly appreciated, thank you in advance!
The claim that $0 \to K \to Y \overset{\pi \circ \rho}{\to} G \to 1$ is a central extension, where $K := \ker(\pi \circ \rho)$ is false in general.
For a counterexample, let $Y$ be nilpotent of class $2$, such as the dihedral group of order $8$, $B := Z(Y)$, $X := Y/B$, $A:=X$, and $G$ trivial. Then $K=Y$, so the extension in question is not central.
It is true however if either $Y$ or $X$ is assumed to be perfect.
Note that $B \le K$ and $K/B \cong A$. We have $[Y,K],Y] = [[K,Y],Y] \le [Y,B] = 1$ so, by the three subgroups lemma, $[[Y,Y],K] = 1$. So if $Y$ is perfect then $[Y,B]=1$ and we are done.
If $X$ is perfect then $Y = [Y,Y]B$, so $[Y,K] = [[Y,Y]B,K] = [[Y,Y],K][B.K] = 1$, so the extension in question is central.