Differential forms in $\mathbb{R} ^2$ and the actuation of $\mathrm{d}$ are as before, but now $f$, $g$, $h$ and $k$ have compact support

$\text{ker }\mathrm{d}_0$ requires again the functions $f$ to be constant, but since compact support is also required, it turns out that $\text{ker }\mathrm{d}_0=\{f\equiv 0\}$ and

For the next group, it was initially proposed the function

that satisfied $\mathrm{d} f=g\mathrm{d} x+h\mathrm{d} y$, so every closed 1-form was exact. The issue is that $f$ may not have compact support even when $g$ and $h$ do. But this is easily fixed: if $g$ and $h$ are zero outside some compact set $[-M,M] \times[-M,M]$, there it is also true that

so $f$ is constant in $\mathbb{R} ^2\smallsetminus [-M,M] \times[-M,M]$ (which is connected). So we may redefine $f$ substracting this constant, so that it is also zero outside $[-M,M] \times[-M,M]$. Therefore, in the compact case every closed 1-form is exact too, and

Here comes the true difference: $H^2_c(\mathbb{R} ^2)$ is not trivial. In the non-compact case, from the 2-form $k\mathrm{d} x\wedge\mathrm{d} y$ we had taken

for $\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}=k$, but now it now it is very rare that $h$ has compact support. Can we choose $g$ and $h$ more wisely? Well, in fact, hardly ever...

The major obstruction is when $\int_{\mathbb{R} ^2} k(x,y)\neq 0$. Imagine that $g$ and $h$ are zero outside some $K=[-M,M] \times[-M,M]$ (and its boundary). Then

so if $\int_{\mathbb{R} ^2} k(x,y)\neq 0$ there's nothing to do: $k\mathrm{d} x\wedge\mathrm{d} y$ is automatically closed, but it's not exact. And what if $\int_{\mathbb{R} ^2} k(x,y)=0$? We'll try the approach above and modify it if needed. Let

Above, below and to the left of $K$, $h$ is identically zero, whereas to the right it equates some function $v(y)$ with support in $[-M,M]$. To correct this anomaly, let's take a differentiable function $\rho(x)$ that values 0 in $(-\infty,-M]$ and 1 in $[M,\infty)$, and we switch to

which now has compact support! But this change comes at a cost that we expect to compensate with $g$:

and we need $\dfrac{\partial g}{\partial y}=-\rho'(x)v(y)$, and we're forced to take

Great! $g$ is zero outside $K$, also above it, because

Since $\int_{\mathbb{R} ^2} k(x,y)$ determines the exactness of the 2-form, there is one degree of freedom and