Chain Rule for Pushforwards

[sanad]

prove that if $g:Y→Z$ and $f:X→Y$ are two smooth maps between a smooth manifolds, then a homomorphism that induced are fulfilling :$(g◦f)∗=f∗◦g∗\, :\, H∙(Z)→H∙(X)$
I must to prove this by a differential forms, but I do not how I can use them .
I began in this way:
if f∗ : H(Y)→H(X), g∗ H(Z)→H(Y) , f∗ H(Y)→H(X),g∗ : H(Z)→H(Y) (by de Rham cohomology) then f∗g∗(ω): H(Z)→H(X), f∗g∗(ω): H(Z)→H(X), now I want to show that (g∘f)∗(ω)=f∗(g∗(ω)),(g∘f)∗(ω)=f∗(g∗(ω)), (g∘f)∗(ω)=ω((g∘f))(g∘f)∗(ω)=ω((g∘f)).
but I do not succeed to finish the proof.

 

[fresh_42]

How would you start? What have you tried so far?
And chain rule is chain rule, so your question comes down to playing around with definitions and equations. You should tell us what you want to use.

 

[fresh_42]

First of all: please embed equations in ## equation ##. Then leave a blank between : and H. The editor otherwise interprets it as an icon. Cp. https://www.physicsforums.com/help/latexhelp/

You have to tell us more about the notations you use and where your letters live in.
In my world I have $(gf)_*(X) = D(gf)(X)=(D(g)(f(X)) D(f)(X)= g_* f_*(X)$ with a vector field $X$ and I am done. So tell us your environment.

 

[WWGD]

Sanad, do you know how to pullback differential forms? We have a map f: X--> Y and an n- form w defined in H_n(Y) , how w is pulled back to H_n(X)? EDIT : This is just multi-linear algebra. Start with a 1-form, i.e., a linear map from X--> Y and find its dual map.

 

[sanad]

@WWGD, can i take the map:# pci:R→S1# for my qeustion"
find an example of #ψ: C•→D•# co-chain map that exists :
#ψi:Ci→Di# is an surjective map (for i>=0) but #ψ∗: (H^k)(C•)→(H^k)D•)# is not a surjective map (for k>=0)? I want to check if an identity map or # pci:R→S1# map can be an example of surjective cochain map like in a question? if not , how I can find like this map? The difintion of cochain map in a booklet is: cochain map between cochain complex #⟨D∙,δ⟩# and #⟨C∙,d⟩# is a homomorphisim chains #ϕn:Cn→Dn#
(for n>=0) such as #δn◦ϕn=ϕn+1◦dn#
"?
Remark: #ψ∗=D(pci)#

 

[fresh_42]

Maybe you want to have a look at the sections for pullbacks and pushforwards in here:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/The Wikipedia entry for de Rham is also worth a read:
https://en.wikipedia.org/wiki/De_Rham_cohomologyor on a specific example:
https://www.physicsforums.com/threads/why-the-terms-exterior-closed-exact.871875/#post-5474443

 

[WWGD]

@WWGD, can i take the map:$pc_i:\mathbb→ R \rightarrow S^1$ for my qeustion"
find an example of $ψ: C•→D•$ co-chain map that exists :
$ψi:Ci→Di$ is an surjective map (for i>=0) but $ψ∗: (H^k)(C•)→(H^k)D•)$ is not a surjective map (for $k\geq 0$)? I want to check if an identity map or $pci:\mathbb R→S^1$map can be an example of surjective cochain map like in a question? if not , how I can find like this map? The difintion of cochain map in a booklet is: cochain map between cochain complex $⟨D∙,δ⟩$ and $⟨C∙,d⟩$ is a homomorphisim chains $ϕn:C_n→D_n$
(for n>=0) such as $δ_n◦ϕ_n=ϕ_n+1◦d_n$
"?
Remark: $ψ∗=D(pc_i)$

Hi Sanad, sorry for the delay. Please use double ##'s at each side to tag Tex.
Are you working with a specific choice of cohomology or you want an argument/example that is independent of the choice of cohomology? Also, I am not sure I understand what the map pci you refer to. Can you state it more explicitly?

 

[WWGD]

This is kind of confusing because differential forms; cohomology in general, pulls back, aka is contravariant , and does not pushforward. Vector fields, under some conditions, push forward
Ok, let's do an example for pullback of a form by a map $\alpha : \mathbb R^n \rightarrow \mathbb R^m$. For definiteness, say $\omega$ is a 2-form and $m,n \geq 3$ ( Aiming for an example that is meaningful but not trivial).

We have:
$\alpha:= (f_1(x_1,...,x_n), f_2(x_1,...,x_n),...,...f_m(x_1,x_2,...,x_n))$ , say

$\omega = w_1dxdy +w_2dxdz+w_3dydz$

Then $df_1 = f_{1x_1}(x_1,x_2,..,x_n)\frac {\partial f_1}{\partial x_1}+...+f_{1 x_n} (x_1,...,x_n) \frac{\partial f_1}{\partial x_n}... df_k =f_{kx_1}(x_1,x_2,...,x_n)\frac {\partial f_k}{\partial x_1}+...+f_{k x_n} (x_1,...,x_n)\frac {\partial f_k}{\partial x_n} ... ..... df_m =f_{mx_1}(x_1,x_2,...,x_n) \frac {\partial f_m}{\partial x_1}+...+f_{m x_n} (x_1,...,x_n) \frac {\partial f_m}{\partial x_n} ...$