I guess that the de Rham cohomology $H^{*}_{dR}(M)$ of a smooth manifold $M$ is naturally isomorphic to a cellular cohomology with coefficient $\mathbb{R}$. To do this, we need only check that de Rham cohomologies setisfy the axioms of generalized cohomology theory since I already know the following theorem:
Theorem. Given an ordinary cohomology theory $H^*(\_,\_;\pi)$ with coefficient $\pi$, then there exists a natural isomorphism $H^*(X,A;\pi)\cong H^*(C^*(X,A;\pi))$ under which the natural transformation $\delta$ agrees with the natural transformation induced by the connecting homomorphism associated to the short exact sequence $$0\rightarrow C^*(X,A;\pi)\rightarrow C^*(X;\pi)\rightarrow C^*(A;\pi)\rightarrow 0.$$
Here $C^*(X,A)$ is a cellular cochain complex of a CW pair $(X,A)$. Furthermore, I will write the axioms of ordinary cohomology theory:
Definition. Let $q$ denote all integers. An ordinary cohomology theory $H^{*}$ with coefficient Abelian group $\pi$ is defined to be a system of functors $H^q(X, A;\pi)$ from the homotopy category of pairs of spaces to the category of Abelian groups and natural transformations $\delta :H^q(A;\pi)\rightarrow H^{q+1}(X,A;\pi)$, where $H^q(X;\pi):=H^q(X, \varnothing;\pi)$, that satisfy the following axioms:
DIMENSION. If $X$ is a point, then for all $q\neq0$, $$H^0(X;\pi)=\pi ;$$ $$H^q(X;\pi)=0.$$ EXACTNESS. The following sequence is exact, where $i:A\rightarrow X$ and $j:(X, \varnothing)\rightarrow (X, A)$ are the inclusions: $$...\rightarrow H^q(X,A;\pi) \xrightarrow{j^*} H^q(X;\pi) \xrightarrow{i^*} H^q(A;\pi) \xrightarrow{\delta} H^{q+1}(X,A;\pi) \rightarrow ...$$ EXCISION. If $(X; A, B)$ is an excisive triad, then the inclusion $(A, A\cap B)\rightarrow (X, B)$ induces an isomorphism $$H^*(X,B;\pi)\rightarrow H^*(A, A\cap B;\pi)$$ ADDITIVITY. If $(X, A)$ is the disjoint union of a set of pairs $(X_i, A_i)$, then the inclusions $(X_i, A_i)\rightarrow (X, A)$ induce an isomorphism $$H^*(X,A;\pi)\rightarrow \prod_i H^*(X_i, A_i;\pi)$$ WEAK EQUIVALENCE. If $f:(X, A)\rightarrow (Y, B)$ is a weak equivalence, then $$f^*:H^*(Y,B:\pi)\rightarrow H^*(X,A;\pi)$$ is an isomorphism.
Let $\Omega^*(M)$ denote the de Rham complex of a smooth manifold $M$. I define the de Rham complex of $M$ together with a smooth embedded submanifold $L$ of $M$ to be
$$\Omega^*(M,L):=ker(\Omega^*(M)\rightarrow \Omega^*(L)),$$
where $\Omega^*(M)\rightarrow \Omega^*(L)$ is a cochain map specified by the pushout map of the smooth embedded inclusion $L\rightarrow M$. In this notation, we define an $n$-th de Rham cohomology of $(M,L)$ to be
$$H^{n}_{dR}(M,L):=H^n(\Omega^*(M,L)).$$
Does the de Rham cohomology theory $H^*_{dR}(\_,\_)$ satisfy all of the axioms? I have already seen the dimension axiom
$$H^0_{dR}(M)=\mathbb{R};$$ $$H^q_{dR}(M)=0$$
if $M$ is a point and $q\neq0$. So I would like to verify the others. But I cannot do it. For example, to do the exactness, I wish that the following sequence of cochain complexes is exact:
$$0\rightarrow\Omega^*(M,L)\rightarrow\Omega^*(M)\rightarrow\Omega^*(L)\rightarrow0$$
because the desired sequence will be obtained by applying cohomology to the above exact sequence. But I cannot find the exactness of $\Omega^*(M)\rightarrow\Omega^*(L)\rightarrow0$. In the other parts of it, I cannot also do a thing. Is my attempt incorrect? Could you give me advice, or some references? I look forward to hearing from you.