$\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.