In $\mathbb{R}^2$, differential forms are as follows:

• 0-forms: $f:\mathbb{R}^2\longrightarrow\mathbb{R}$ differentiable
• 1-forms: $g\mathrm{d} x+h\mathrm{d} y$, $g$, $h:\mathbb{R}^2\longrightarrow\mathbb{R}$ differentiable
• 2-forms: $k\mathrm{d} x\wedge\mathrm{d} y$, $k:\mathbb{R}^2\longrightarrow\mathbb{R}$ differentiable

and the exterior derivative acts on them:

• $\mathrm{d}(f)=\mathrm{d} f=\dfrac{\partial f}{\partial x}\mathrm{d} x+\dfrac{\partial f}{\partial y}\mathrm{d} y$
• $\mathrm{d}(g\mathrm{d} x+h\mathrm{d} y)= \mathrm{d} g\wedge\mathrm{d} x+\mathrm{d} h\wedge\mathrm{d} y= \left(\dfrac{\partial g}{\partial x}\mathrm{d} x+\dfrac{\partial g}{\partial y}\mathrm{d} y\right)\wedge\mathrm{d} x+\left(\dfrac{\partial h}{\partial x}\mathrm{d} x+\dfrac{\partial h}{\partial y}\mathrm{d} y\right)\wedge\mathrm{d} y\\ \phantom{\mathrm{d}(g\mathrm{d} x+h\mathrm{d} y)}= \dfrac{\partial g}{\partial x}\mathrm{d} x\wedge\mathrm{d} x+\dfrac{\partial g}{\partial y}\mathrm{d} y\wedge\mathrm{d} x+\dfrac{\partial h}{\partial x}\mathrm{d} x\wedge\mathrm{d} y+\dfrac{\partial h}{\partial y}\mathrm{d} y\wedge\mathrm{d} y= \left(\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}\right)\mathrm{d} x\wedge\mathrm{d} y$
• $\mathrm{d}(k\mathrm{d} x\wedge\mathrm{d} y)= \mathrm{d} k\wedge\mathrm{d} x\wedge\mathrm{d} y= \left(\dfrac{\partial k}{\partial x}\mathrm{d} x+\dfrac{\partial k}{\partial y}\mathrm{d} y\right)\wedge\mathrm{d} x\wedge\mathrm{d} y= \dfrac{\partial k}{\partial x}\mathrm{d} x\wedge\mathrm{d} x\wedge\mathrm{d} y+\dfrac{\partial k}{\partial y}\mathrm{d} y\wedge\mathrm{d} x\wedge\mathrm{d} y=0$

Now we may compute the cohomology groups

• $\text{ker }\mathrm{d}_0=\left\{f\middle|\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial y}=0\right\}=\{f|f=cte\in\mathbb{R}\}$
• $\text{im }\mathrm{d}_{-1}=\{0\}$ (we take $\mathrm{d}_{-1}:0\longrightarrow\Omega^0(\mathbb{R}^2)$ the zero map)

• $\text{ker }\mathrm{d}_1=\left\{g\mathrm{d} x+h\mathrm{d} y\middle|\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}=0\right\}$
• $\text{im }\mathrm{d}_0=\left\{\dfrac{\partial f}{\partial x}\mathrm{d} x+\dfrac{\partial f}{\partial y}\mathrm{d} y\right\}$

Obviously, $\text{im }\mathrm{d}_0\subset\text{ker }\mathrm{d}_1$, because $\dfrac{\partial (\partial f/\partial y)}{\partial x}-\dfrac{\partial (\partial f/\partial x)}{\partial y}=0$. Now we wonder whether equality holds. Let $g$, $h$ with $\dfrac{\partial h}{\partial x}=\dfrac{\partial g}{\partial y}$; there will exist some $f$ with $\dfrac{\partial f}{\partial x}=g$, $\dfrac{\partial f}{\partial y}=h$? If so, this function could be found by integration, moving along the $x$-axis and then vertically; let $f(0,0)=0$:

Does this function meet our requirements?

so $f$ is our desired function. Therefore there is equality in our previous sets and

• $\text{ker }\mathrm{d}_2=\Omega^2(\mathbb{R}^2)=\{k\mathrm{d} x\wedge\mathrm{d} y\}$
• $\text{im }\mathrm{d}_1=\left\{\left(\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}\right)\mathrm{d} x\wedge\mathrm{d} y\right\}$

Given some fixed function $k$, there will exist $g$, $h$ with $\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}=k$? Now is very easy: we may take $g=0$ and $h(x,y)=\int_0^x k(t,y)\mathrm{d} t$, and everything is ok. So again kernel and image coincide and