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I am following Ararat Babakhanian's Cohomological Methods in Group theory.

Let $A,B,C$ be $G$ modules then we have a $G$ module structre on $\text{Hom}_{\mathbb{Z}}(B,C)$ with $$\sigma.f(x)=\sigma(f\sigma^{-1}x)$$

$ A\times B\xrightarrow{\cdot } C$ is called a $G$ pairing if $\sigma a\cdot \sigma b=\sigma (a\cdot b)$ for all $\sigma\in G$ and $a\in A, b\in B$.

By Universal property of tensor products, this induces a unique linear map $\Omega : A\otimes B\rightarrow C$ such that $\Omega (a\otimes b)=a\cdot b$.

There is then a homomorphism of cohomology groups $H^{p+q}(G,A\otimes B)\xrightarrow{\Omega^*} H^{p+q}(G,C)$ induced from $\Omega : A\otimes B\rightarrow C$.

We have cup product map $H^p(G,A)\times H^q(G,B)\xrightarrow{\cup} H^{p+q}(G,A\otimes B)$, composing with $\Omega^*$ we have the map $$H^p(G,A)\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

He then says we have evaluation (Bilinear) map $$\text{Hom}_{\mathbb{Z}}(B,C)\times B\rightarrow C$$

with $(f,b)\mapsto f(b)$. I checked that this is a $G$ pairing.

In this case we should get map

$$H^p(G,\text{Hom}_{\mathbb{Z}}(B,C) )\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

But then, he writes there is a map

$$H^p(G,\text{Hom}(B,C) )\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

Where $\text{Hom}(B,C)$ denote all $G$ module homomorphisms..

So, what happened here? Though We can restrict evaluation mp to $\text{Hom}(B,C)$, there is no module strutre on this, so we can not talk about cohomology or anything..

This continued in further material.. Let me know what is the point here.

enter image description here

enter image description here

I am following Ararat Babakhanian's Cohomological Methods in Group theory.

Let $A,B,C$ be $G$ modules then we have a $G$ module structre on $\text{Hom}_{\mathbb{Z}}(B,C)$ with $$\sigma.f(x)=\sigma(f\sigma^{-1}x)$$

$ A\times B\xrightarrow{\cdot } C$ is called a $G$ pairing if $\sigma a\cdot \sigma b=\sigma (a\cdot b)$ for all $\sigma\in G$ and $a\in A, b\in B$.

By Universal property of tensor products, this induces a unique linear map $\Omega : A\otimes B\rightarrow C$ such that $\Omega (a\otimes b)=a\cdot b$.

There is then a homomorphism of cohomology groups $H^{p+q}(G,A\otimes B)\xrightarrow{\Omega^*} H^{p+q}(G,C)$ induced from $\Omega : A\otimes B\rightarrow C$.

We have cup product map $H^p(G,A)\times H^q(G,B)\xrightarrow{\cup} H^{p+q}(G,A\otimes B)$, composing with $\Omega^*$ we have the map $$H^p(G,A)\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

He then says we have evaluation (Bilinear) map $$\text{Hom}_{\mathbb{Z}}(B,C)\times B\rightarrow C$$

with $(f,b)\mapsto f(b)$. I checked that this is a $G$ pairing.

In this case we should get map

$$H^p(G,\text{Hom}_{\mathbb{Z}}(B,C) )\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

But then, he writes there is a map

$$H^p(G,\text{Hom}(B,C) )\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

Where $\text{Hom}(B,C)$ denote all $G$ module homomorphisms..

So, what happened here? Though We can restrict evaluation mp to $\text{Hom}(B,C)$, there is no module strutre on this, so we can not talk about cohomology or anything..

This continued in further material.. Let me know what is the point here.

I am following Ararat Babakhanian's Cohomological Methods in Group theory.

Let $A,B,C$ be $G$ modules then we have a $G$ module structre on $\text{Hom}_{\mathbb{Z}}(B,C)$ with $$\sigma.f(x)=\sigma(f\sigma^{-1}x)$$

$ A\times B\xrightarrow{\cdot } C$ is called a $G$ pairing if $\sigma a\cdot \sigma b=\sigma (a\cdot b)$ for all $\sigma\in G$ and $a\in A, b\in B$.

By Universal property of tensor products, this induces a unique linear map $\Omega : A\otimes B\rightarrow C$ such that $\Omega (a\otimes b)=a\cdot b$.

There is then a homomorphism of cohomology groups $H^{p+q}(G,A\otimes B)\xrightarrow{\Omega^*} H^{p+q}(G,C)$ induced from $\Omega : A\otimes B\rightarrow C$.

We have cup product map $H^p(G,A)\times H^q(G,B)\xrightarrow{\cup} H^{p+q}(G,A\otimes B)$, composing with $\Omega^*$ we have the map $$H^p(G,A)\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

He then says we have evaluation (Bilinear) map $$\text{Hom}_{\mathbb{Z}}(B,C)\times B\rightarrow C$$

with $(f,b)\mapsto f(b)$. I checked that this is a $G$ pairing.

In this case we should get map

$$H^p(G,\text{Hom}_{\mathbb{Z}}(B,C) )\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

But then, he writes there is a map

$$H^p(G,\text{Hom}(B,C) )\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

Where $\text{Hom}(B,C)$ denote all $G$ module homomorphisms..

So, what happened here? Though We can restrict evaluation mp to $\text{Hom}(B,C)$, there is no module strutre on this, so we can not talk about cohomology or anything..

This continued in further material.. Let me know what is the point here.

enter image description here

enter image description here

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user37663
user37663

Pairing in Group Cohomology

I am following Ararat Babakhanian's Cohomological Methods in Group theory.

Let $A,B,C$ be $G$ modules then we have a $G$ module structre on $\text{Hom}_{\mathbb{Z}}(B,C)$ with $$\sigma.f(x)=\sigma(f\sigma^{-1}x)$$

$ A\times B\xrightarrow{\cdot } C$ is called a $G$ pairing if $\sigma a\cdot \sigma b=\sigma (a\cdot b)$ for all $\sigma\in G$ and $a\in A, b\in B$.

By Universal property of tensor products, this induces a unique linear map $\Omega : A\otimes B\rightarrow C$ such that $\Omega (a\otimes b)=a\cdot b$.

There is then a homomorphism of cohomology groups $H^{p+q}(G,A\otimes B)\xrightarrow{\Omega^*} H^{p+q}(G,C)$ induced from $\Omega : A\otimes B\rightarrow C$.

We have cup product map $H^p(G,A)\times H^q(G,B)\xrightarrow{\cup} H^{p+q}(G,A\otimes B)$, composing with $\Omega^*$ we have the map $$H^p(G,A)\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

He then says we have evaluation (Bilinear) map $$\text{Hom}_{\mathbb{Z}}(B,C)\times B\rightarrow C$$

with $(f,b)\mapsto f(b)$. I checked that this is a $G$ pairing.

In this case we should get map

$$H^p(G,\text{Hom}_{\mathbb{Z}}(B,C) )\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

But then, he writes there is a map

$$H^p(G,\text{Hom}(B,C) )\times H^q(G,B)\rightarrow H^{p+q}(G,C)$$

Where $\text{Hom}(B,C)$ denote all $G$ module homomorphisms..

So, what happened here? Though We can restrict evaluation mp to $\text{Hom}(B,C)$, there is no module strutre on this, so we can not talk about cohomology or anything..

This continued in further material.. Let me know what is the point here.