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If you want to find another $\sigma'_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'_{U,V\otimes W}=\sigma'_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'_{V,W}\sigma'_{W,V}=Id_{V\otimes W}$, instantly shows that $f(1,1)=\pm1$. But of course there were many assumptions made along the way.

As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^*(M)$ (over a field), you then can apply the K"unneth formula and say $H^*(M\times N)=H^*(M)\otimes H^*(N)$. In what sense these two are isomorphic as algebras? To define aa product on the tensor product $H^*(M)\otimes H^*(N)$, you need those isomorphisms $\sigma$, so that you can do $H^*(M)\otimes H^*(N)\otimes H^*(M)\otimes H^*(N)\to H^*(M)\otimes H^*(M)\otimes H^*(N)\otimes H^*(N)$ and then compute the product in $H^*(M)$ and in $H^*(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.

If you want to find another $\sigma'_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'_{U,V\otimes W}=\sigma'_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'_{V,W}\sigma'_{W,V}=Id_{V\otimes W}$, instantly shows that $f(1,1)=\pm1$. But of course there were many assumptions made along the way.

As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^*(M)$ (over a field), you then can apply the Künneth formula and say $H^*(M\times N)=H^*(M)\otimes H^*(N)$. In what sense these two are isomorphic as algebras? To define aa product on the tensor product $H^*(M)\otimes H^*(N)$, you need those isomorphisms $\sigma$, so that you can do $H^*(M)\otimes H^*(N)\otimes H^*(M)\otimes H^*(N)\to H^*(M)\otimes H^*(M)\otimes H^*(N)\otimes H^*(N)$ and then compute the product in $H^*(M)$ and in $H^*(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.

If you want to find another $\sigma'\_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'\_{U,V\otimes W}=\sigma'\_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'\_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'(V,W)\sigma'(W,V)=Id\_{V\otimes W}$, instantly shows that $f(1,1)=\pm 1$. But of course there were many assumptions made along the way.

As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^\ast(M)$ (over a field), you then can apply the K"unneth formula and say $H^\ast(M\times N)=H^\ast(M)\otimes H^\ast(N)$. In what sense these two are isomorphic as algebras? To define aa product on the tensor product $H^\ast(M)\otimes H^\ast(N)$, you need those isomorphisms $\sigma$, so that you can do $H^\ast(M)\otimes H^\ast(N)\otimes H^\ast(M)\otimes H^\ast(N)\to H^\ast(M)\otimes H^\ast(M)\otimes H^\ast(N)\otimes H^\ast(N)$ and then compute the product in $H^\ast(M)$ and in $H^\ast(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.

If you want to find another $\sigma'_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'_{U,V\otimes W}=\sigma'_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'_{V,W}\sigma'_{W,V}=Id_{V\otimes W}$, instantly shows that $f(1,1)=\pm1$. But of course there were many assumptions made along the way.

As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^*(M)$ (over a field), you then can apply the K"unneth formula and say $H^*(M\times N)=H^*(M)\otimes H^*(N)$. In what sense these two are isomorphic as algebras? To define aa product on the tensor product $H^*(M)\otimes H^*(N)$, you need those isomorphisms $\sigma$, so that you can do $H^*(M)\otimes H^*(N)\otimes H^*(M)\otimes H^*(N)\to H^*(M)\otimes H^*(M)\otimes H^*(N)\otimes H^*(N)$ and then compute the product in $H^*(M)$ and in $H^*(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.

If you want to find another $\sigma'\_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'\_{U,V\otimes W}=\sigma'\_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'\_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'(V,W)\sigma'(W,V)=Id\_{V\otimes W}$, instantly shows that $f(1,1)=\pm 1$. But of course there were many assumptions made along the way.

As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^\*(M)$ (over a field), you then can apply the K"unneth formula and say $H^\*(M\times N)=H^\*(M)\otimes H^\*(N)$. In what sense these two are isomorphic as algebras? To define a product on the tensor product $H^\*(M)\otimes H^\*(N)$, you need those isomorphisms $\sigma$, so that you can do $H^\*(M)\otimes H^\*(N)\otimes H^\*(M)\otimes H^\*(N)\to H^\*(M)\otimes H^\*(M)\otimes H^\*(N)\otimes H^\*(N)$ and then compute the product in $H^\*(M)$ and in $H^\*(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.

If you want to find another $\sigma'\_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'\_{U,V\otimes W}=\sigma'\_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'\_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'(V,W)\sigma'(W,V)=Id\_{V\otimes W}$, instantly shows that $f(1,1)=\pm 1$. But of course there were many assumptions made along the way.

As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^\ast(M)$ (over a field), you then can apply the K"unneth formula and say $H^\ast(M\times N)=H^\ast(M)\otimes H^\ast(N)$. In what sense these two are isomorphic as algebras? To define aa product on the tensor product $H^\ast(M)\otimes H^\ast(N)$, you need those isomorphisms $\sigma$, so that you can do $H^\ast(M)\otimes H^\ast(N)\otimes H^\ast(M)\otimes H^\ast(N)\to H^\ast(M)\otimes H^\ast(M)\otimes H^\ast(N)\otimes H^\ast(N)$ and then compute the product in $H^\ast(M)$ and in $H^\ast(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.