$\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.