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Let $M$ and $N$ be manifolds. If $f$, $g:M\longrightarrow N$ are homotopic, the maps $f^\ast$, $g^\ast:H^k(N)\longrightarrow H^k(M)$ coincide

Before dealing with the main statement, let's study the forms in the space $M\times \mathbb{R} $.

In $\mathbb{R} ^{n+1}$, a basis for the alternate forms of order $k$ is $\{\mathrm{d} x_{i_1}\wedge...\wedge\mathrm{d} x_{i_k}:1\leqslant i_1 < ... < i_k \leqslant n+1\}$. Some of this elements have $\mathrm{d} x_{n+1}$ but others don't; dividing into two groups, any $k$-form is decomposable in a unique way as $\beta+\alpha\wedge\mathrm{d} x_{n+1}$, where $\beta$ is a $k$-form in $\mathbb{R} ^n$ and $\alpha$ is a $(k-1)$-form in $\mathbb{R} ^n$

When considering manifolds, any $k$-form in $M\times \mathbb{R} $ is decomposable in a unique way as $\beta+\mathrm{d} t\wedge\alpha$, where $\beta$ and $\alpha$ are forms in $M$ of order $k$ and $k-1$ dependent on $x\in M$ and $t\in \mathbb{R} $. That is, given a chart of $M$ these forms are expressed as

$$\beta=\sum \beta_{i_1 i_2... i_k}(x,t)\mathrm{d} x_{i_1}\wedge...\wedge\mathrm{d} x_{i_k},\qquad 1\leqslant i_1 < ... < i_k \leqslant n$$ $$\alpha=\sum \alpha_{i_1 i_2... i_{k-1}}(x,t)\mathrm{d} x_{i_1}\wedge...\wedge\mathrm{d} x_{i_{k-1}},\qquad 1\leqslant i_1 < ... < i_{k-1} \leqslant n$$

Usual calculus in the second argument allows us to derive and integrate a form $\beta(t)\in \Omega^k(M)$ along the $\mathbb{R}$ fiber:

$$ \begin{array}{rl} \displaystyle\dfrac{\partial \beta}{\partial t}, & \displaystyle\qquad\dfrac{\partial \beta}{\partial t}(v_1,...,v_k)=\dfrac{\partial }{\partial t}[\beta(v_1,...,v_k)]\\ \displaystyle\int_a^b \beta, & \displaystyle\qquad\left(\int_a^b \beta\right)(v_1,...,v_k)=\int_a^b (\beta(v_1,...,v_k))\\ \end{array} $$

The first operation outputs a $k$-form in $M$ for each $t$ in which derivation is performed; the second operation outputs a $k$-form for each interval $[a,b]$

Let $\mathrm{d}$ be the exterior derivative in $M$ and $\bar{\mathrm{d}}$ the exterior derivative in $M\times \mathbb{R} $. For brevity, $\mathrm{d} x_{i_1}\wedge...\wedge\mathrm{d} x_{i_k}$ will be denoted $\mathrm{d} x_I$

$$\bar{\mathrm{d}}(f(x,t)\mathrm{d} x_I)=\bar{\mathrm{d}}(f(x,t))\wedge\mathrm{d} x_I=\left(\sum \dfrac{\partial f}{\partial x_i}\mathrm{d} x_i + \dfrac{\partial f}{\partial t}\mathrm{d} t\right)\wedge\mathrm{d} x_I=\mathrm{d}(f\mathrm{d} x_I)+\mathrm{d} t\wedge\left(\dfrac{\partial f}{\partial t}\mathrm{d} x_I\right)$$ $$\bar{\mathrm{d}}(f(x,t)\mathrm{d} t\wedge\mathrm{d} x_I)=\bar{\mathrm{d}}(f(x,t))\wedge \mathrm{d} t \wedge\mathrm{d} x_I=\left(\sum \dfrac{\partial f}{\partial x_i}\mathrm{d} x_i + \dfrac{\partial f}{\partial t}\mathrm{d} t\right)\wedge\mathrm{d} t\wedge\mathrm{d} x_I=-\sum \dfrac{\partial f}{\partial x_i}\mathrm{d} t\wedge\mathrm{d} x_i\wedge\mathrm{d} x_I=-\mathrm{d} t\wedge\mathrm{d}(f(x,t)\wedge\mathrm{d} x_I)$$

and summing up everything,

$$\bar{\mathrm{d}}(\beta+\mathrm{d} t\wedge\alpha)=\mathrm{d} \beta+\mathrm{d} t\wedge\left(\dfrac{\partial \beta}{\partial t}-\mathrm{d} \alpha\right)$$

Now we return to the main problem. We have $f$, $g:M\longrightarrow N$ inducing $f^\ast$, $g^\ast:\Omega^k(N)\longrightarrow \Omega^k(M)$. And we'd like to build a homotopy of complex homomorphisms between $f^\ast$ and $g^\ast$, much in the way the prism operator provided a homotopy for homology. What are our tools? We have a homotopy $$H:M\times [0,1]\longrightarrow N$$ $$h_t:M\longrightarrow N,\qquad h_t(x)=H(x,t)$$ $$h_0=f\qquad h_1=g$$ $$h_t=H\circ s_t$$

with $s_t$ the natural immersion of $M$ in $M\times\mathbb{R} $ at level $t$. That is,

$$\Omega^k(N)\xrightarrow{h_t^\ast}\Omega^k(M)\quad\leadsto\quad\Omega^k(N)\xrightarrow{H^\ast}\Omega^k(M\times\mathbb{R} )\xrightarrow{s_t^\ast}\Omega^k(M)$$

Let's do it! Let $\gamma\in\Omega^k(N)$, with $H^\ast(\gamma)=\beta+\mathrm{d} t\wedge\alpha\in\Omega^k(M\times\mathbb{R} )$. In the inclusion $s_t$, the vertical component does nothing, so $$s_t^\ast(\beta+\mathrm{d} t\wedge\alpha)=\beta(t)$$

This way,

$$f^\ast(\gamma)=(H\circ s_0)^\ast(\gamma)=s_0^\ast(H^\ast(\gamma))=s_0^\ast(\beta+\mathrm{d} t\wedge\alpha)=\beta(0)$$ $$g^\ast(\gamma)=(H\circ s_1)^\ast(\gamma)=s_1^\ast(H^\ast(\gamma))=s_1^\ast(\beta+\mathrm{d} t\wedge\alpha)=\beta(1)$$

We define the following operator

$$K:\Omega^k(N)\longrightarrow\Omega^{k-1}(M),\qquad K(\gamma)=\int_0^1\alpha$$

Is this a homotopy between $f^\ast$ and $g^\ast$? Does $\mathrm{d}\circ K+K\circ\mathrm{d}=g^\ast-f^\ast$ hold? On one side

$$\mathrm{d} K(\gamma)=\mathrm{d} \int_0^1\alpha=\int_0^1\mathrm{d}\alpha$$

because integration behaves as a ponderated sum of the $\alpha$ found along the fiber, and $\mathrm{d}$ is a linear operator. On the other side,

$$H^\ast(\mathrm{d} \gamma)=\bar{\mathrm{d}} H^\ast(\gamma)=\bar{\mathrm{d}}(\beta+\mathrm{d} t\wedge\alpha)=\mathrm{d} \beta+\mathrm{d} t\wedge\left(\dfrac{\partial \beta}{\partial t}-\mathrm{d} \alpha\right)$$

and therefore

$$K\mathrm{d} \gamma=\int_0^1\left(\dfrac{\partial \beta}{\partial t}-\mathrm{d} \alpha\right)=\beta(1)-\beta(0)-\int_0^1\mathrm{d}\alpha$$

having finally

$$(\mathrm{d} K+K\mathrm{d})(\gamma)=\int_0^1\mathrm{d}\alpha+\beta(1)-\beta(0)-\int_0^1\mathrm{d}\alpha=\beta(1)-\beta(0)=(g^\ast -f^\ast)(\gamma)$$

so $K$ effectively induces a homotopy between $f^\ast$ and $g^\ast$ and the maps in cohomology coincide. It is remarkable that the $K$ operator is indeed closely related to the prism operator: if $\sigma$ is a submanifold of $M$ whose dimension matches the order of $\gamma$, then