If what you said is: $U(1) \cong \mathbb R \setminus \mathbb Z \cong \S^{1} $ Why not try to calculate the $H^3[U(1)\times U(1),U(1)] \cong H^d[]$ in that way as a product? In that case, you can simply use the cellular cohomology where the cohomology groups are directly obtained via applying the $Hom(_,Z)$ onto the corresponding homology chain complex $H_{k}(U(1)\times U(1),U(1)])$.

Step1: Observe or calculate the $H_{d}(U(1)\times U(1),U(1)])$ via Device Lemma

Lemma 2.34 in A.Hatcher's Algebraic Topology If X is a CW complex, then: (a) $H_{k}(Xn,Xn−1)$ is zero for k!=n and is free abelian for k=n, with a basis in one-to-one correspondence with the n-cells of X. (b) $H_{k}(X^{n})=0$ for k > n. In particular, if X is finite-dimensional then $H_{k}(X)=0$ for k > $dimX$. (c) The inclusion i :Xn->X induces an isomorphism I: $H_{k}(X^{n})->H_{k}(X)$ if k < n.

Step2: Calculate the cohomology via applying the $Hom(_,Z)$

Ans1:I do not think the results are correct for Q1.

I don't know if there's any restriction by doing this through cellular cohomology in PHYSIC background.

For further ref. you can search by the keyword"Eilenberg-MacLane space" to deal with the infinite dimensional case.

Again,I strongly suggest that you can just switch the subscripts to suit the cohomology... You're just using the UCThm for homology in the notations for UCThm of cohomology, which makes it confusing...Moreover, I'm not sure whether you're talking about external product.

Ans2So it's meaningless to talk about 'inconsistency' unless you correctly present the Q2 arguments.

If what you said is: $U(1) \cong \mathbb R \setminus \mathbb Z \cong \S^{1} $ Why not try to calculate the $H^3[U(1)\times U(1),U(1)] \cong H^d[S^{1} \times S^{1},S^{1}]$ in that way as a product? In that case, you can simply use the cellular cohomology where the cohomology groups are directly obtained via applying the $Hom( ,Z)$ onto the corresponding homology chain complex $H_{k}[U(1)\times U(1),U(1)]$.

Step1: Observe or calculate the $H_{k}[U(1)\times U(1),U(1)]$ via Device Lemma

Lemma 2.34 in A.Hatcher's Algebraic Topology If X is a CW complex, then: (a) $H_{k}(Xn,Xn−1)$ is zero for k!=n and is free abelian for k=n, with a basis in one-to-one correspondence with the n-cells of X. (b) $H_{k}(X^{n})=0$ for k > n. In particular, if X is finite-dimensional then $H_{k}(X)=0$ for k > $dimX$. (c) The inclusion i :Xn->X induces an isomorphism I: $H_{k}(X^{n})->H_{k}(X)$ if k < n.

Step2: Calculate the cohomology via applying the $Hom( ,Z)$

Ans1:I do not think the results are correct for Q1.

I don't know if there's any restriction by doing this through cellular cohomology in PHYSIC background.

For further ref. you can search by the keyword"Eilenberg-MacLane space" to deal with the infinite dimensional case.

Again,I strongly suggest that you can just switch the subscripts to suit the cohomology... You're just using the UCThm for homology in the notations for UCThm of cohomology, which makes it confusing...Moreover, I'm not sure whether you're talking about external product.

Ans2:So it's meaningless to talk about 'inconsistency' unless you correctly present the Q2 arguments.

Plz treat this as a comment: If what you said is: $U(1) \cong \mathbb R \setminus \mathbb Z \cong \wedge^{\mathbb Z} S^{1} $ Why not try to calculate the $H^3[U(1)\times U(1),U(1)]$ in that way as a wedge product?

I strongly suggest that you can just switch the subscripts to suit the cohomology... You're just using the UCThm for homology in the notations for UCThm of cohomology, which makes it confusing...

If what you said is: $U(1) \cong \mathbb R \setminus \mathbb Z \cong \S^{1} $ Why not try to calculate the $H^3[U(1)\times U(1),U(1)] \cong H^d[]$ in that way as a product? In that case, you can simply use the cellular cohomology where the cohomology groups are directly obtained via applying the $Hom(_,Z)$ onto the corresponding homology chain complex $H_{k}(U(1)\times U(1),U(1)])$.

Step1: Observe or calculate the $H_{d}(U(1)\times U(1),U(1)])$ via Device Lemma

Lemma 2.34 in A.Hatcher's Algebraic Topology If X is a CW complex, then: (a) $H_{k}(Xn,Xn−1)$ is zero for k!=n and is free abelian for k=n, with a basis in one-to-one correspondence with the n-cells of X. (b) $H_{k}(X^{n})=0$ for k > n. In particular, if X is finite-dimensional then $H_{k}(X)=0$ for k > $dimX$. (c) The inclusion i :Xn->X induces an isomorphism I: $H_{k}(X^{n})->H_{k}(X)$ if k < n.

Step2: Calculate the cohomology via applying the $Hom(_,Z)$

Ans1:I do not think the results are correct for Q1.

I don't know if there's any restriction by doing this through cellular cohomology in PHYSIC background.

For further ref. you can search by the keyword"Eilenberg-MacLane space" to deal with the infinite dimensional case.

Again,I strongly suggest that you can just switch the subscripts to suit the cohomology... You're just using the UCThm for homology in the notations for UCThm of cohomology, which makes it confusing...Moreover, I'm not sure whether you're talking about external product.

Ans2So it's meaningless to talk about 'inconsistency' unless you correctly present the Q2 arguments.