Is Deligne's braiding functorial? $\newcommand{\ssc}{{\rm sc}}
\newcommand{\ad}{{\rm ad}}
\newcommand{\Fbar}{{\overline F}}
$
Let $F$ be a field and $\Fbar$ be a fixed algebraic closure of $F$.
Let $G$ be a (connected) reductive group over $F$.
Let $G^\ssc$ denote the universal cover of the commutator subgroup $[G,G]$ of $G$.
Following Deligne, Variétés de Shimura: interprétation modulaire et techniques de construction de modèles canoniques, Proc. Sympos. Pure Math 33, Part 2, 1979, Section 2.0.11, we consider the composite homomorphism
$$\rho\colon\ G^\ssc\to [G,G]\to G.$$
Deligne (loc. cit., Section 2.0.2)  noticed that the commutator map
$$[\ ,]\colon\ G\times G\to G,\quad\ g_1,g_2\mapsto [g_1,g_2] := g_1 g_2 g_1^{-1} g_2^{-1}$$
lifts to a certain map (morphism of $F$-varieties)
$$ \lbrace\ , \rbrace \colon\ G\times G\to G^\ssc,\quad\ g_1,g_2\mapsto \lbrace g_1,g_2 \rbrace$$
as follows.
The commutator map
$$G^\ssc\times G^\ssc\to G^\ssc,\quad\ s_1,s_2\mapsto [s_1,s_2]:= s_1 s_2 s_1^{-1} s_2^{-1}$$
clearly factors via  a morphism of $F$-varieties
$$(G^\ssc)^\ad\times (G^\ssc)^\ad\to G^\ssc$$
where $(G^\ssc)^\ad=G^\ssc/Z_{G^\ssc}$ and $Z_{G^\ssc}$ denotes the center of $G^\ssc$.
Identifying $(G^\ssc)^\ad$ with $G^\ad:= G/Z_G$, we obtain the desired morphism of $F$-varieties
$$\lbrace\ ,\rbrace\colon\ G\times G\to G^\ad\times G^\ad\to G^\ssc.$$
On $\Fbar$-points, if $g_1,g_2\in G(\Fbar),\ g_1=\rho(s_1) z_1,\ g_2=\rho(s_2) z_2$
where $s_1,s_2\in G^\ssc(\Fbar),\ z_1,z_2\in Z_G(\Fbar)$, then
$$ \lbrace g_1,g_2\rbrace=[s_1,s_2].$$
The constructed map $\lbrace\ ,\rbrace$ has nice properties, in particular,
$$ \rho\big(\lbrace g_1,g_2\rbrace\big)=[g_1,g_2]\qquad\text{and}
\qquad \lbrace g_1,g_2\rbrace=\lbrace g_2,g_1\rbrace^{-1}.$$
Actually,  $\lbrace\ ,\rbrace$ is a  symmetric braiding
of the crossed module $(G^\ssc\to G)$.
We call it Deligne's braiding.
Now let $\varphi\colon G\to H$ be a homomorphism of reductive $F$-groups.
It induces a homomorphism $\varphi^\ssc\colon G^\ssc\to H^\ssc$.
The maps
$$ [\ ,]\colon\ G\times G\to G,\ g_1,g_2\mapsto [g_1,g_2]\quad\text{and}
  \quad [\ ,]\colon\ G^\ssc\times G^\ssc\to G^\ssc,\ s_1,s_2\mapsto [s_1,s_2]$$
are functorial in $G$:
$$ \varphi\big([g_1,g_2]\big)=\big[\varphi(g_1),\varphi(g_2)\big]\quad\text{and}
   \quad  \varphi^\ssc\big([s_1,s_2]\big)=\big[\varphi^\ssc(s_1),\varphi^\ssc(s_2)\big].$$

Question. Is Deligne's braiding functorial?
In other words, is it true that for any homomorphism $\varphi\colon G\to H$,
we have
$$\varphi^\ssc\big(\lbrace g_1,g_2\rbrace\big)=
\big\lbrace\varphi(g_1),\varphi(g_2)\big\rbrace\quad 
\text{for all}\ \ g_1,g_2 \in G\ ?$$

The answer is Yes when homomorphism  $\varphi$ is normal, that is, $\varphi(G)$ is normal in $H$.
Indeed, then $\varphi$ induces homomorphisms
$$Z_G\to Z_H,\quad Z_{G^\ssc}\to  Z_{H^\ssc},\quad G^\ad\to H^\ad.$$
In general I expect the answer No, but cannot construct a counter-example.
 A: $\newcommand{\ssc}{{\rm sc}}
\newcommand{\sss}{{\rm ss}}
\newcommand{\ad}{{\rm ad}}
\newcommand{\wh}{\widehat}
\newcommand{\wt}{\widetilde}
\newcommand{\pitil}{\tilde\pi}
\newcommand{\rhotil}{\tilde\rho} 
\newcommand{\gtil}{\tilde g}
\newcommand{\stil}{\tilde s}
$The answer is Yes.

Proposition.
Let $\varphi\colon G\to H$ be a homomorphism of connected reductive groups
over an algebraically closed field $F$.
Let $\varphi^\ssc\colon G^\ssc\to H^\ssc$ denote the induced homomorphism.
Then for any $g_1,g_2\in G(F)$ we have
$$ \big\lbrace \varphi(g_1),\varphi_(g_2)\big\rbrace _H=\varphi^\ssc\big\lbrace g_1,g_2\big\rbrace _G $$
where for simplicity we write
$\varphi^\ssc\big\lbrace g_1,g_2\big\rbrace _G$ instead of $\varphi^\ssc\big(\lbrace g_1,g_2\rbrace _G\big)$.

Proof. Since the map $\big\lbrace \ ,\big\rbrace _H$ factors via $H^\ad$, we may and shall assume that $H=H^\ad$.
Consider the homomorphisms $\varphi\colon G\to H$ and $\rho_H\colon H^\ssc\to H$.
The fiber product
$$ \wh G=G\times_H H^\ssc $$
is endowed with two homomorphisms
$$ \hat \pi_G\colon \wh G\to G\quad\text{and}\quad \hat\pi_H\colon\wh G\to H^\ssc.$$
Since the homomorphism $\rho_H\colon H^\ssc\to H$ is surjective with finite kernel,
so is the homomorphism $\hat \pi_G\colon \wh G\to G$.
Let $\wt G$ denote the identity component of $\wh G$.
Let
$$\pitil_G\colon\wt G\to G\quad\text{and}\quad \pitil_H\colon \wt G\to H^\ssc$$
denote the restrictions to $\wt G$ of  $\hat \pi_G$ and $\hat \pi_H$, respectively.
Then $\pitil_G$ is a surjective homomorphism with finite kernel.
It follows that $\wt G$ is a connected reductive $F$-group.
Write $G=C\cdot G^\ssc$ where $C$ is the radical (largest central torus) of $G$,
and $G^\sss=[G,G]$ is the commutator subgroup of $G$.
Similarly, write $\wt G=\wt C\cdot \wt G^\ssc$ where $\wt C$ is the radical of $\wt G$,
and $\wt G^\sss=[\wt G,\wt G]$.
Then we have surjective homomorphisms $\pitil^\sss\colon \wt G^\sss\to G^\sss$
and $\pitil_C\colon \wt C\to C$ with finite kernels.
It follows that there exists a unique surjective homomorphism with finite kernel
$\rhotil\colon G^\ssc\to\wt G^\sss$ such that
$$ \pitil^\sss\circ\rhotil=\rho_G\colon\  G^\ssc\to G^\sss.$$
From the commutative diagram
$\require{AMScd}$
\begin{CD}
G^\ssc    @>\rhotil>>    \wt G^\sss   @>\pitil_H>>  H^\ssc\\
@|              @VV\pitil^\sss V            @VV\rho_H V\\
C^\ssc    @>\rho_G>>    G^\sss       @>\varphi>>  H
\end{CD}
we see that
$$ \varphi^\ssc=\pitil_H\circ \rhotil\colon\ G^\ssc\to H^\ssc.$$
Now let $g_i\in G(F)$, $i=1,2$.
Write
$$ g_i=c_i\cdot s_i \qquad\text{where}\ \ c_i\in C(F),\ s_i\in G^\sss(F). $$
We lift $c_i$ to some $\tilde c_i\in\wt C(F)$ and $s_i$ to some $\stil_i\in \wt G^\sss(F)$.
We set $\gtil_i=\tilde c_i\cdot \stil_i\in \wt G (F)$; then $\pitil_G(\gtil_i)=g_i$.
We lift $\stil_i$ to some $s^\ssc_i\in G^\ssc(F)$.
Set $h_i^\ssc=\pitil_H(\gtil_i)\in H^\ssc(F)$.
Then $\rho_H(h_i^\ssc)=\varphi(g_i)$, and we have
\begin{align*}
 \big\lbrace \varphi(g_1),\varphi(g_2)\big\rbrace _H :&=[h_1^\ssc,h_2^\ssc]=
  \pitil_H[\gtil_1,\gtil_2]\\
 &=\pitil_H[\stil_1,\stil_2]=
  \pitil_H\rhotil [s_1^\ssc,s_2^\ssc]=\varphi^\ssc[s_1^\ssc,s_2^\ssc].
\end{align*}
Since $g_i=c_i\cdot s_i=c_i\cdot\pitil_H(\stil_i)=c_i\cdot \rho_G(s_i^\ssc)$, we see that
$$ \big\lbrace g_1,g_2\big\rbrace _G=[s_1^\ssc,s_2^\ssc]\in G^\ssc(F)$$
and $\varphi^\ssc[s_1^\ssc,s_2^\ssc]=\varphi^\ssc\big\lbrace g_1,g_2\big\rbrace _G.$
Thus
$$ \big\lbrace \varphi(g_1),\varphi(g_2)\big\rbrace _H=\varphi^\ssc\big\lbrace g_1,g_2\big\rbrace _G,$$
as required.
