= Solution
Use an <SU(2)> <connection one-form> $A=A_\mu dx^\mu$ with <Skew-Hermitian> generators $T_a=-i\sigma_a/2$, so $\operatorname{tr}(T_aT_b)=-\delta_{ab}/2$. Absorb the gauge coupling into $A$, and write its <gauge curvature> as $F=dA+A\wedge A$. The <Yang-Mills gauge transformation> convention is $A^h=h^{-1}Ah+h^{-1}dh$, giving $F^h=h^{-1}Fh$.
In four dimensions the <Second Chern form> and <Second Chern number> of the associated fundamental rank-two bundle are
$$
c_2(A)=\frac1{8\pi^2}\operatorname{tr}(F\wedge F),\qquad
k_C=\int_Mc_2(A)\in\mathbb Z.
$$
Here $M$ is a closed oriented four-manifold. This sign follows by expanding the total <Chern class> $\det(1+iF/(2\pi))$ and using $\operatorname{tr}F=0$. The first Chern class vanishes for <SU(2)>; the second is the degree-four characteristic class relevant here. With the same curvature convention the commonly positive self-dual <instanton number> is
$$
Q=-\frac1{8\pi^2}\int_M\operatorname{tr}(F\wedge F)=-k_C.
$$
Both sign conventions occur; fixing them at the outset avoids identifying opposite-oriented integers.
The form is <gauge-invariant> by conjugation and cyclicity of the <matrix trace>. The <gauge-theory Bianchi identity> $D_AF=0$ makes it closed. To see that its integral is independent of the connection, vary $A$:
$$
\delta F=D_A\delta A,\qquad
\delta\operatorname{tr}(F\wedge F)
=2d\,\operatorname{tr}(\delta A\wedge F).
$$
On a closed $M$ the integral of this exact form vanishes. This is <Chern-Weil connection transgression>, showing that $k_C$ depends on the bundle's topology rather than on a particular <gauge field>.
Locally the same four-form is an exterior derivative. The <Chern-Simons 3-form>
$$
\omega_3(A)=\operatorname{tr}\left(A\wedge dA+
\frac23A\wedge A\wedge A\right)
$$
satisfies
$$
d\omega_3=\operatorname{tr}(F\wedge F).
$$
For verification, differentiating gives $\operatorname{tr}(dA\wedge dA+2dA\wedge A\wedge A)$; expansion of $F\wedge F$ gives these terms and $\operatorname{tr}A^4$. Graded cyclicity moves the first one-form through the other three, changing the sign, so $\operatorname{tr}A^4=0$. This local exactness does not make the <Second Chern number> zero on every closed manifold: on a nontrivial bundle the gauge potentials and $\omega_3$ are only patchwise defined.
For the usual finite-action sector on $\mathbb R^4$, impose pure-gauge behavior $A\to h^{-1}dh$ at infinity. Compactification gives a bundle over $S^4$, with transition map $h:S^3\to SU(2)$. The <Maurer-Cartan equation> gives $d(h^{-1}dh)=-(h^{-1}dh)^2$, hence
$$
\omega_3(h^{-1}dh)=-\frac13\operatorname{tr}(h^{-1}dh)^3.
$$
With the outward boundary orientation, <Stokes theorem> therefore gives
$$
k_C=-\frac1{24\pi^2}\int_{S^3_\infty}
\operatorname{tr}(h^{-1}dh)^3,\qquad
Q=\frac1{24\pi^2}\int_{S^3_\infty}
\operatorname{tr}(h^{-1}dh)^3.
$$
These are opposite choices of the <winding number>, and either becomes the <topological degree> of $h$ after specifying the orientation of the target <SU(2)> sphere. Thus integrality also follows from $\pi_3(SU(2))=\mathbb Z$. Smooth deformations preserving the asymptotic sector cannot change it.
The positive Euclidean <Yang-Mills action> is
$$
S=-\frac1{g^2}\int_M\operatorname{tr}(F\wedge *F)
=\frac1{4g^2}\int_MF_{\mu\nu}^aF_{\mu\nu}^a\,d^4x.
$$
Define $\|X\|^2=-\int\operatorname{tr}(X\wedge *X)$. Since the <Hodge star> has square one on two-forms in Euclidean dimension four,
$$
S=\frac1{2g^2}\|F-s*F\|^2+\frac{8\pi^2s}{g^2}Q,
\qquad s=\pm1.
$$
Choosing $s=\operatorname{sgn}Q$ gives the <Yang-Mills instanton Bogomolny bound>
$$
\boxed{S\geq\frac{8\pi^2}{g^2}|Q|,\qquad
F=s*F\ \text{at equality}.}
$$
The <self-dual Yang-Mills equations> and their anti-self-dual version imply the <Yang-Mills equations> $D_A*F=0$ by the <Bianchi identity>. Their finite-action solutions are <Yang-Mills instantons>.
For a concrete self-dual example, the <BPST instanton> with center $a$ and size $\rho>0$ has
$$
A_\mu^a=\frac{2\eta^a_{\mu\nu}(x-a)^\nu}{|x-a|^2+\rho^2},
\qquad
F_{\mu\nu}^a=-\frac{4\rho^2\eta^a_{\mu\nu}}
{(|x-a|^2+\rho^2)^2},
$$
where $\eta^a_{\mu\nu}$ are self-dual <'t Hooft symbols>. Their contraction is $\sum_{a,\mu,\nu}(\eta^a_{\mu\nu})^2=12$. Thus $F_{\mu\nu}^aF_{\mu\nu}^a=192\rho^4/(|x-a|^2+\rho^2)^4$, and
$$
\int_{\mathbb R^4}\frac{\rho^4\,d^4x}{(|x|^2+\rho^2)^4}
=2\pi^2\int_0^\infty\frac{\rho^4r^3\,dr}{(r^2+\rho^2)^4}
=\frac{\pi^2}{6}.
$$
This gives $S=8\pi^2/g^2$ and $Q=1$. Reversing duality gives $Q=-1$. The free center, size and framed gauge orientation illustrate the <instanton moduli space>.
On a three-dimensional slice $\Sigma$, there is no four-form Chern integral intrinsic to the slice. Its transgression is the <Chern-Simons number of a gauge field>, in the instanton-charge convention
$$
N_{\mathrm{CS}}[A]=-\frac1{8\pi^2}\int_\Sigma\omega_3(A).
$$
It is a real number for a general connection, not necessarily an integer, and is not strictly gauge invariant. Set $\theta=h^{-1}dh$ and $\bar\theta=dh\,h^{-1}$. Substituting $A^h=h^{-1}Ah+\theta$, using $d\theta=-\theta^2$ and graded trace cyclicity, gives the <gauge change of the Chern-Simons three-form>
$$
\omega_3(A^h)=\omega_3(A)-d\operatorname{tr}(\bar\theta\wedge A)
-\frac13\operatorname{tr}\theta^3.
$$
The mixed terms collect into the displayed exact form; the last term is also fixed by the pure-gauge case $A=0$. Therefore on a closed $\Sigma$, or with boundary conditions making the exact-form integral vanish,
$$
\boxed{N_{\mathrm{CS}}[A^h]=N_{\mathrm{CS}}[A]+w(h),\qquad
w(h)=\frac1{24\pi^2}\int_\Sigma\operatorname{tr}(h^{-1}dh)^3\in\mathbb Z.}
$$
Thus $N_{\mathrm{CS}}\bmod\mathbb Z$, or $\exp(2\pi iN_{\mathrm{CS}})$, is gauge invariant. Gauge maps homotopic to the identity have $w=0$; maps of nonzero winding are large transformations even when they equal the identity at infinity.
For a <Yang-Mills vacuum> on $\mathbb R^3$, zero magnetic <energy> requires $F_{ij}=0$. On simply connected space this is a pure gauge $A=h^{-1}dh$, and $N_{\mathrm{CS}}=w(h)$ is an integer after fixing the asymptotic trivialization. Quotienting only by transformations homotopic to the identity leaves integer-labelled vacuum representatives. Allowing all winding transformations identifies them classically; quantum states can instead transform by a phase, producing the <theta vacuum>. A three-dimensional pure-gauge representative carries a winding number but is not a localized positive-energy monopole. In fact <no static finite-energy lump in three-dimensional pure Yang-Mills theory> exists with the usual decay conditions: $A_i^{(s)}(x)=sA_i(sx)$ gives $E(s)=sE(1)$, and stationarity forces $E=0$.
Finally, on an oriented spacetime slab $[t_-,t_+]\times\Sigma$, with no side-boundary contribution, the transgression identity gives
$$
\boxed{Q=N_{\mathrm{CS}}(t_+)-N_{\mathrm{CS}}(t_-).}
$$
A <Yang-Mills instanton> therefore interpolates between vacuum representatives whose <Chern-Simons numbers> differ by its integer charge. In a four-dimensional quantum theory a <Yang-Mills theta term> weights such sectors by $e^{i\theta Q}$, making $\theta$ periodic modulo $2\pi$. In an intrinsically three-dimensional theory an action $2\pi\ell N_{\mathrm{CS}}$ has a gauge-invariant phase $e^{iS}$ only for integer level $\ell$ under all winding gauge maps. These statements distinguish the gauge-invariant integer four-dimensional Chern charge from the three-dimensional connection-dependent quantity defined modulo integers.
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