= Solution
Give $\mathfrak g$ its <Adjoint representation> and use the <tensor product of Lie algebra representations>. Define the <action map of a Lie algebra representation>
$$
\boxed{a:\mathfrak g\otimes V\longrightarrow V,\qquad a(x\otimes v)=xv.}
$$
For $y\in\mathfrak g$, the action on the source is $y(x\otimes v)=[y,x]\otimes v+x\otimes yv$. The defining representation identity then gives
$$
a(y(x\otimes v))=[y,x]v+x(yv)=y(xv)=y\,a(x\otimes v).
$$
Thus $a$ is a <Lie algebra representation homomorphism>.
If $V$ is nontrivial and simple, the action is not identically zero, so the image of $a$ is a nonzero submodule. Simplicity forces $a$ to be surjective. For a finite-dimensional $V$ and a <semisimple Lie algebra>, the <Weyl complete reducibility theorem> provides an invariant complement $U$ to $\ker a$ in $\mathfrak g\otimes V$. The restriction $a|_U$ is then an isomorphism onto $V$, giving
$$
\boxed{\mathfrak g\otimes V\cong\ker a\oplus V.}
$$
This proves the requested summand assertion in the finite-dimensional setting.
If finite dimensionality is not implicit, the final assertion needs that hypothesis: the <action summand can fail for an infinite-dimensional simple module>. Here is an explicit counterexample. Take $\mathfrak g=\mathfrak{sl}_2$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$, and the <Verma module> of <highest weight> $-2$:
$$
V=\bigoplus_{m\ge0}\mathbb Cv_m,\qquad
fv_m=v_{m+1},\quad hv_m=(-2-2m)v_m,\quad ev_m=-m(m+1)v_{m-1}.
$$
The lowering coefficient never vanishes for $m>0$, so an <invariant subspace>, using the distinct $h$-weights to isolate a basis vector and then raising, contains $v_0$ and hence all of $V$. Thus $V$ is simple.
In $\mathfrak{sl}_2\otimes V$, a vector of <weight> $-2$ is $A e\otimes v_1+B h\otimes v_0$. Applying $e$ gives $-2(A+B)e\otimes v_0$, so its highest vectors are precisely the multiples of $w=e\otimes v_1-h\otimes v_0$. But $w=f(e\otimes v_0)$, and $e\otimes v_0$ has <weight> zero, absent from $V$. Every homomorphism $r:\mathfrak g\otimes V\to V$ therefore kills this vector and then kills $w$. Any embedded copy of $V$ would send its highest vector to a nonzero multiple of $w$, so it cannot admit a projection back to $V$. This rules out a direct summand, not merely a splitting of the particular action map.
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