The de rham cohomology class of a form is defined without closedness

Given: The smooth one-form $\alpha =xdy$ on ${\mathbb{R}}^2$.

Closed and exact differential forms: For the complex def-de-rham-cochain-complex, put $Z^k(M)=\ker (d:{\mathrm{\Omega }}^k(M)\to {\mathrm{\Omega }}^{k+1}(M))$ and $B^k(M)=\mathrm{im}(d:{\mathrm{\Omega }}^{k-1}(M)\to {\mathrm{\Omega }}^k(M))$. A form is closed if it belongs to $Z^k$ and exact if it belongs to $B^k$. If $\omega =d\eta$, then $d\omega =d^2\eta =0$ by thm-the-exterior-derivative-squares-to-zero, so $B^k\subseteq Z^k$. In particular $B^0=0$, since ${\mathrm{\Omega }}^{-1}=0$. The zero form is both closed and exact in every degree.

De rham cohomology: The real de Rham cohomology is $H_{\mathrm{d}\mathrm{R}}^k(M)=Z^k(M)\mathrm{/}{B}^k(M)$, with $Z^k,B^k$ as in def-closed-and-exact-differential-forms. This is def-cohomology-object-of-a-cochain-complex in real vector spaces. Only a closed form $\omega$ represents a class $[\omega ]$. For closed forms $\omega ,\omega \mathrm{\prime }$, equality $[\omega ]=[\omega \mathrm{\prime }]$ means precisely $\omega \mathrm{\prime }-\omega =d\eta$ for some $(k-1)$-form $\eta$. Addition and real scalar multiplication are induced by those of forms. All groups on the empty manifold are zero.

The local coordinate formula for the exterior derivative: Let $(U,x^1,\dots ,x^n)$ be a smooth chart on a smooth manifold and $\omega$ a smooth $k$-form on $U$, with $k\ge 0$. Summing over increasing $k$-tuples $I$, and writing $d{x}^I=d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$, if $\omega ={\sum }_I{\omega }_{I}d{x}^I$, then $d\omega ={\sum }_{I}d{\omega }_I\wedge d{x}^I\mathrm{.}$