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Note that not all elements of $(U \otimes V) \otimes W$ are simple tensors. This is the problem you are encountering, as your map isn't defined for elements like $(u_1 \otimes v_1) \otimes w_1 + (u_2 \otimes v_2) \otimes w_2$. You can fix this by mapping each simple tensor as above and then extending the map linearly on all elements. This trivially gets ...

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In general the tensor product does not preserve injective maps. For example, consider $A=\mathbb Z/2\mathbb Z$, $C=\mathbb Z$ and $B=2\mathbb Z$. Then the inclusion induces the zero map $$\mathbb Z/2\mathbb Z\otimes 2\mathbb Z\to\mathbb Z/2\mathbb Z\otimes \mathbb Z$$ since $$\overline a\otimes 2b\mapsto\overline a\otimes 2b=\overline{2a}\otimes b=0\otimes b=... 2 For finite dimensional algebras, flat modules are projective, so the projective dimension of M is the same as its weak dimension, which is$$\sup\{d\mid \text{Tor}^A_d(M,-)\neq0\}.$$Since the class of left modules X such that \text{Tor}^A_d(M,X)=0 is closed under coproducts and extensions, and every module is an iterated extension of coproducts of ... 2 For a finite-dimensional vector space V, V is isomorphic to its double dual V^{**}, through the natural isomorphism \varphi:V\rightarrow V^{**} given by \varphi(x)(v)=v(x), where x\in V and v\in V^*. Hence, an element of V can be seen as a linear functional that acts on covectors. 2 That is correct (for \alpha an element of the base ring). However, using that as a definition hides the most important property of the tensor product, namely the universal property that it satisfies. 2 u and v are members of some vector space V over a field F, and \phi and \theta are members of the corresponding dual vector space V^*. So T is a linear map T:V \times V \times V^* \times V^* \rightarrow F such that T(u,v,\phi,\theta) = \phi(u)\theta(v) Since T is a (2,2) tensor, we can write it in component form as T^{ab}_{cd}. ... 2 If you like a more algebraic approach, consider the map$$ f : \mathbb Z \to \mathbb Z / m \mathbb Z \otimes_{\mathbb Z} \mathbb Z / n \mathbb Z,\; a \mapsto a \otimes 1. $$This is clearly surjective, so we must have$$ \mathbb Z / m \mathbb Z \otimes_{\mathbb Z} \mathbb Z / n \mathbb Z \simeq \mathbb Z / \ker f. $$Now all that is left is to show that \... 1 Yes, it's the same, in a sense. The property you seek is the existence of a canonical isomorphism between spaces L(X_1,X_2)\otimes L(Y_1,Y_2) and L(X_1\otimes Y_1,X_2\otimes Y_2), for X_1,X_2,Y_1,Y_2 being arbitrary vector spaces. This isomorphism$$ \iota : L(X_1,X_2)\otimes L(Y_1,Y_2) \rightarrow L(X_1\otimes Y_1,X_2\otimes Y_2)$$can be defined by ... 1 As usual, we represent ket \newcommand{\ket}{\lvert{#1}\rangle}\newcommand{\bra}{\langle{#1}\rvert} \ket{a_1} by a column vector \mathbb{C}^N=\mathbb{C}^{N\times 1} and bra \bra{a_2} by a row vector \mathbb{C}^N=\mathbb{C}^{1\times N}. The outer product of two kets \ket{a_1} and \ket{a_2} is the ket-bra matrix product \ket{a_1}\bra{a_2}\... 1 If with respect to a coordinate system (x^j) on M you write J(\partial_j) = \sum J^i_{~j}\partial_i, then J can be regarded as$$J = \sum J^i_{~j}\,{\rm d}x^j \otimes \partial_i,$$acting on vector fields by$$X \mapsto J(X) = \sum J^i_{~j}{\rm d}x^j(X)\partial_i.$$There is nothing particular about manifolds or almost-complex structures here. In ... 1 It means that any polynomial z(s,t) can be represented as a sum of polynomials, each of the form x(s)y(t). 1 The indices in the coefficients come from the objects where you evaluate the tensor on. Since g is covariant, it eats vectors. If (e_1,e_2) are dual to (\epsilon^1, \epsilon^2), you can evaluate g_{ij} = g(e_i,e_j), but not g^{ij} = g(\epsilon^1,\epsilon^2), as the latter a priori does not make sense. Then$$\begin{align} g_{11} &= g(e_1,e_1) = ...

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It probably cant get better than the author of the article answering himself but here a small addition that also answers the question and gives a direct interpretation of this Tor, namely it counts something. Since we do homological algebra, we can assume that the algebra is basic. As in Jeremy Rickards answer one has $Tor_d^A(M,A/radA)=DExt_A^d(M,D(A/radA))$...

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One short way of getting the desired isomorphism is \begin{align*} (K \otimes A) \otimes_K (K \otimes B) &\cong (A \otimes K) \otimes_K (K \otimes B) \\ &\cong A \otimes (K \otimes_K (K \otimes B)) \tag{$\ast$} \\ &\cong A \otimes K \otimes B \\ &\cong K \otimes A \otimes B \\ &\cong K \otimes (A \otimes B) ...

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The correct symbol $\otimes$ is produced with \otimes. A strategy is to first find the matrix of $T$ taken with respect to the basis $\mathcal{B}$, and then convert it to the standard basis. With respect to $\mathcal{B}$, it is clear that $$[T]_{\mathcal{B}} = \begin{pmatrix} 0 & 1 \\-1 & 2 \end{pmatrix}.$$Now, we use the tensor transformation law (...

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You probably meant $\otimes$ instead of $\oplus$. Try using the fact that the matrix elements of a tensor $M=a\otimes b$ in the canonical basis are $M_{ij}=v_i\cdot Mv_j = (a\cdot v_i)(b\cdot v_j)$ where $\cdot$ is the standard dot product and $v_k$ are the basis vectors.

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