[proofplan] We construct an explicit chain homotopy between the pullback maps $f_0^*$ and $f_1^*$ on the de Rham complexes. The operator is obtained by pulling a form on $N$ back to $[0,1]\times M$, inserting the time-direction vector field, and integrating the resulting family of forms over $[0,1]$. The chain homotopy identity shows that the difference $f_1^*\omega-f_0^*\omega$ is exact whenever $\omega$ is closed, so the two pullbacks induce the same map on cohomology. [/proofplan]

[step:Define the homotopy operator by integrating contraction in the time direction] Since $f_0$ and $f_1$ are smoothly homotopic, choose a smooth homotopy \begin{align*} F: [0,1]\times M &\to N \end{align*} from $f_0$ to $f_1$, where $[0,1]$ is regarded as a smooth manifold with boundary. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $[0,1]$. Thus, for each $t\in[0,1]$, the time-slice map \begin{align*} f_t: M &\to N \\ p &\mapsto F(t,p) \end{align*} is smooth, with $f_t=F\circ\iota_t$ for the inclusion \begin{align*} \iota_t: M &\to [0,1]\times M \\ p &\mapsto (t,p). \end{align*} In particular $f_0 = F\circ \iota_0$ and $f_1 = F\circ \iota_1$. Let $T \in \mathfrak{X}([0,1]\times M)$ denote the smooth vector field tangent to the $[0,1]$ factor, defined by \begin{align*} T_{(t,p)}=\frac{\partial}{\partial t}\bigg|_{(t,p)}. \end{align*} For each integer $k\geq 1$, define a [linear map](/page/Linear%20Map) \begin{align*} I_k: \Omega^k([0,1]\times M) \to \Omega^{k-1}(M) \end{align*} as follows. If $\eta \in \Omega^k([0,1]\times M)$, $p \in M$, and $v_1,\dots,v_{k-1}\in T_pM$, then \begin{align*} (I_k\eta)_p(v_1,\dots,v_{k-1}) = \int_0^1 \eta_{(t,p)} \bigl( T_{(t,p)}, (d\iota_t)_p(v_1), \dots, (d\iota_t)_p(v_{k-1}) \bigr) \, d\mathcal{L}^1(t). \end{align*} For $k=0$, set $I_0=0$, where $\Omega^{-1}(M)$ is interpreted as the zero [vector space](/page/Vector%20Space). Now define, for every integer $k\geq 0$, the homotopy operator \begin{align*} K_k: \Omega^k(N) \to \Omega^{k-1}(M) \end{align*} by \begin{align*} K_k\omega = I_k(F^*\omega). \end{align*} When $k=0$, this means $K_0=0$. [/step]

[step:Prove the chain homotopy identity on the product cylinder]We claim that for every integer $k\geq 0$ and every $\eta \in \Omega^k([0,1]\times M)$, \begin{align*} \iota_1^*\eta-\iota_0^*\eta = d(I_k\eta)+I_{k+1}(d\eta). \end{align*} Indeed, write the unique product decomposition \begin{align*} \eta=\alpha_t+dt\wedge\beta_t, \end{align*} where $\alpha_t\in\Omega^k(M)$ and $\beta_t\in\Omega^{k-1}(M)$ are smooth families of forms on $M$. Then \begin{align*} I_k\eta=\int_0^1\beta_t\,d\mathcal{L}^1(t). \end{align*} Using the product formula for the [exterior derivative](/page/Exterior%20Derivative), \begin{align*} d\eta = d_M\alpha_t + dt\wedge \left( \frac{\partial \alpha_t}{\partial t} - d_M\beta_t \right). \end{align*} Therefore \begin{align*} d(I_k\eta)+I_{k+1}(d\eta) &= \int_0^1 d_M\beta_t\,d\mathcal{L}^1(t) + \int_0^1 \left( \frac{\partial \alpha_t}{\partial t} - d_M\beta_t \right) \,d\mathcal{L}^1(t) \\ &= \int_0^1 \frac{\partial \alpha_t}{\partial t} \,d\mathcal{L}^1(t) \\ &= \alpha_1-\alpha_0. \end{align*} Since $\iota_t^*dt=0$, we have $\iota_t^*\eta=\alpha_t$, and hence \begin{align*} d(I_k\eta)+I_{k+1}(d\eta) = \iota_1^*\eta-\iota_0^*\eta. \end{align*}[/step]

[guided]The point of this step is to isolate the calculation on the cylinder $[0,1]\times M$. Once it is proved there, applying it to $\eta=F^*\omega$ will give the desired formula for $f_1^*\omega-f_0^*\omega$. We use the product coordinate $t: [0,1]\times M \to [0,1]$ and write $dt$ for its differential. Every $k$-form $\eta\in\Omega^k([0,1]\times M)$ decomposes uniquely as \begin{align*} \eta=\alpha_t+dt\wedge \beta_t, \end{align*} where, for each $t\in[0,1]$, $\alpha_t\in\Omega^k(M)$ and $\beta_t\in\Omega^{k-1}(M)$ depend smoothly on $t$. In this notation, \begin{align*} I_k\eta=\int_0^1 \beta_t\,d\mathcal{L}^1(t), \end{align*} meaning that for $p\in M$ and $v_1,\dots,v_{k-1}\in T_pM$, \begin{align*} (I_k\eta)_p(v_1,\dots,v_{k-1}) = \int_0^1 (\beta_t)_p(v_1,\dots,v_{k-1}) \,d\mathcal{L}^1(t). \end{align*} The [exterior derivative](/page/Exterior%20Derivative) on the product separates into the derivative in the $M$ variables and the derivative in the $t$ variable by the product formula for differential forms. Therefore \begin{align*} d\eta &= d_M\alpha_t + dt\wedge \frac{\partial \alpha_t}{\partial t} - dt\wedge d_M\beta_t \\ &= d_M\alpha_t + dt\wedge \left( \frac{\partial \alpha_t}{\partial t} - d_M\beta_t \right), \end{align*} where $d_M$ denotes the [exterior derivative](/theorems/1525) on $M$ applied at fixed $t$. Hence \begin{align*} I_{k+1}(d\eta) = \int_0^1 \left( \frac{\partial \alpha_t}{\partial t} - d_M\beta_t \right) \,d\mathcal{L}^1(t). \end{align*} Also, since [exterior differentiation](/page/Exterior%20Derivative) on $M$ is computed coefficientwise in local coordinates, it commutes with integration over the compact interval $[0,1]$ of the smooth family of forms $\beta_t$: \begin{align*} d(I_k\eta) = d_M\left(\int_0^1 \beta_t\,d\mathcal{L}^1(t)\right) = \int_0^1 d_M\beta_t\,d\mathcal{L}^1(t). \end{align*} Adding the two identities gives \begin{align*} d(I_k\eta)+I_{k+1}(d\eta) &= \int_0^1 \frac{\partial \alpha_t}{\partial t} \,d\mathcal{L}^1(t) \\ &= \alpha_1-\alpha_0. \end{align*} The last equality is the one-dimensional [Fundamental Theorem of Calculus](/theorems/632) applied coefficientwise to the smooth family of forms $\alpha_t$. Since $\iota_t^*dt=0$ and $\iota_t^*\alpha_t=\alpha_t$, we have \begin{align*} \iota_t^*\eta=\alpha_t. \end{align*} Thus \begin{align*} d(I_k\eta)+I_{k+1}(d\eta) = \iota_1^*\eta-\iota_0^*\eta. \end{align*} This proves the chain homotopy identity.[/guided]

[step:Apply the chain homotopy identity to the pulled back form] Let $k\geq 0$ and let $\omega\in\Omega^k(N)$. Apply the identity from the previous step to the form $\eta=F^*\omega\in\Omega^k([0,1]\times M)$. Since [exterior differentiation](/page/Exterior%20Derivative) commutes with [pullback](/page/Pullback) by a smooth map, \begin{align*} d(F^*\omega)=F^*(d\omega). \end{align*} Also, \begin{align*} \iota_1^*(F^*\omega) = (F\circ\iota_1)^*\omega = f_1^*\omega, \end{align*} and \begin{align*} \iota_0^*(F^*\omega) = (F\circ\iota_0)^*\omega = f_0^*\omega. \end{align*} Substituting into the cylinder identity gives \begin{align*} f_1^*\omega-f_0^*\omega &= d(I_k(F^*\omega))+I_{k+1}(d(F^*\omega)) \\ &= d(K_k\omega)+K_{k+1}(d\omega). \end{align*} Thus the pullback chain maps $f_0^*$ and $f_1^*$ are chain homotopic. [/step]

[step:Pass from the chain homotopy identity to de Rham cohomology] Let $k\geq 0$ and let $[\omega]\in H_{\mathrm{dR}}^k(N)$ be a de Rham cohomology class represented by a closed form $\omega\in\Omega^k(N)$, so $d\omega=0$. We use the convention already implicit in $K_0=0$ that $\Omega^{-1}(M)$ is the zero vector space and that the differential out of this zero space is the zero map. The identity from the previous step gives \begin{align*} f_1^*\omega-f_0^*\omega = d(K_k\omega)+K_{k+1}(d\omega) = d(K_k\omega). \end{align*} Thus $f_1^*\omega-f_0^*\omega$ is exact on $M$. Therefore the two closed forms $f_0^*\omega$ and $f_1^*\omega$ represent the same cohomology class in $H_{\mathrm{dR}}^k(M)$: \begin{align*} [f_1^*\omega]=[f_0^*\omega]. \end{align*} Since this holds for every closed representative $\omega\in\Omega^k(N)$ and every integer $k\geq 0$, the induced maps on de Rham cohomology agree: \begin{align*} f_0^*=f_1^*:H_{\mathrm{dR}}^k(N)\to H_{\mathrm{dR}}^k(M). \end{align*} [/step]