### CAlg tensor products, sync-pdf tool

parent 11e366a1
 ... ... @@ -172,9 +172,11 @@ Packages for sectioning and positioning. \DeclareMathOperator{\ev}{ev} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\kernel}{Ker} \DeclareMathOperator{\kernel}{ker} \DeclareMathOperator{\coker}{coker} \DeclareMathOperator{\Ann}{Ann} \DeclareMathOperator{\Image}{Im} \DeclareMathOperator{\image}{im} \DeclareMathOperator{\spn}{span} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\rang}{rang} ... ... @@ -192,6 +194,7 @@ Packages for sectioning and positioning. \DeclareMathOperator{\ggT}{ggT} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\hht}{ht} \DeclareMathOperator{\Alg}{-Alg} % Cat ... ...
 ... ... @@ -31,7 +31,7 @@ Let $R$ be a ring. Then \end{itemize} \end{proof} \begin{rem}[] \begin{rem}[]\label{rem:units-in-local} Let $R$ be a ring and $I \subseteq R$ an ideal. Then \begin{align*} R^{\times} = R - I \iff R \text{ is local and } I = m_R ... ...
 ... ... @@ -7,3 +7,32 @@ We define and construct tensor products for modules over a ring, and then for al \subsection{Definition/Construction} \input{./files/05/05a-construction} \subsection{Functoriality of the tensor product} If clear from the context, we will omit the ring $R$ in the notation $M \otimes_R N$ and instead write $M \otimes N$ and just write linear instead of $R$-linear. \input{./files/05/05b-functoriality} \subsection{Tensor product and direct sums} \input{./files/05/05c-direct-sum} \subsection{Commutativity and associativity} \input{./files/05/05d-commutativity-associativity} \subsection{Exactness} \input{./files/05/05e-exactness} \subsection{Base Change} \input{./files/05/05f-base-change} \subsection{Tensor product of algebras} \input{./files/05/05g-algebras} \subsection{Multilinear Algebra (mostly without proofs)} \input{./files/05/05h-multilinear} \subsection{Tensor, Symmetric and Alternating Algebras} \input{./files/05/05i-symmetric-alternating}
 ... ... @@ -85,6 +85,10 @@ Before we give the definition, we make a small detour to the category $\Set$\foo \end{rem} \textbf{Notation\footnote{This is neither standard nor used by the lecturer, but we will sometimes use the Lambda-like notation for function application for higher-order functions. So for $f \in \Hom(X,\Hom(Y,Z))$, we write $f\ x$ for $f(x) \in \Hom(Y,Z)$ and $f\ x \ y$ for $(f(x))(y) \in Z$}}: \begin{thm}[Whitney] Let $M,N$ be $R$-modules. Then there exists an $R$-module $M \otimes_R N$ ($M$ tensor $N$ over $R$) and a bilinear map ... ... @@ -144,4 +148,67 @@ In other words, there is a natural equivalence We construct $M \otimes N$ as a quotient of a huge'' free module with a submodule that imposes the bilinearity by applying the $\Hom$ properties of free modules aswell as those of quotient modules. Let $\tilde{T}$ be the free $R$-module with basis induced by the elements of $M \times N$ and write $[(m,n)]$ for the basis vector corresponding to $(m,n)$. Let $Z \subseteq \tilde{T}$ be the submodule generaated by the elements \begin{align*} [(rm,n)] - r[(m,n)], \quad [(m,rn)] - r[(m,n)]\\ [(m+m',n)] - [(m,n)] - [(m',n)], \quad [(m',n+n')] - [(m,n)] - [(m,n')] \end{align*} now define \begin{align*} M \otimes_R N = \faktor{\tilde{T}}{Z}, \quad \beta(m,n) = \pi([(m,n)]), \quad \text{where} \quad \pi: \tilde{T} \to \faktor{\tilde{T}}{Z} \end{align*} This pair $(M \otimes_R N,\beta)$ satisfies the conditions we want. Let $P$ be an $R$-module. By the characterisation of the hom-functor for quotient maps, we have \begin{align*} \Hom_R(M \otimes N,P) \stackrel{\iso}{\to} \left\{ g: \tilde{T} \to P \big\vert Z \subseteq \kernel g \right\}, \quad f \mapsto f \circ \pi \end{align*} By the characterisation of free modules, we have a bijection \begin{align*} \Hom_R(\tilde{T},P) \iso \Hom_{\Set}(M \times N,P) \end{align*} Therefore, af $f \in \Hom_R(M \otimes_R N,P)$ corresponds to a function \begin{align*} h: M \times N \to P \quad \text{such that} \quad h)(m,n) = f(\pi([m,n])) \end{align*} which is exactly bilinearity of $h$. We conclude \begin{align*} \Hom_R(M \otimes_R N,P) \iso \Bil_R(M,N;P) \end{align*} \end{proof} \begin{rem}[] Since $\beta$ is bilinear, the elements in $M \otimes_R N$ satisfy \begin{align*} rm \otimes n = r(m \otimes n) = m \otimes rn\\ (m + m') \otimes n = m \otimes n + m' \otimes n\\ m \otimes (n+n') = m \otimes n + m \otimes n' \end{align*} \textbf{Warning:} Not all elements of $M \otimes_R N$ are \emph{pure tensors} of the form $m \otimes n$, but instead finite linear combinations \begin{align*} \sum_{i=1}^{n}m_i \otimes n_i, \quad n \geq 0, m_i \in M, n_i \in N \end{align*} so the collection of elements $m \otimes n$ generate $M \otimes_R N$. \end{rem} It is useful to know that \begin{align*} 0 \otimes n = (0 \cdot 0) \otimes n = 0 \cdot (0 \otimes n) = 0 = m \otimes 0 \end{align*} \begin{xmp}[] If we know that there cannot be any non-trivial bilinear maps in $\Bil_R(M,N;P)$, then we know that $M \otimes_R N = \{0\}$. For example, let $m,n$ be coprime integers. Then \begin{align*} \Z/m\Z \otimes_\Z/n\Z = \{0\} \quad \text{because} \quad 1 = am + bn, \text{ for } a,b \in \Z \end{align*} \end{xmp}
 Given two $R$-modules $M,N$ we have constructed a new $R$-module $M \otimes_R N$. If we fix $M$, we obtain functors $M \otimes_R -, - \otimes_R M: R_{\text{Mod}} \to R_{\text{Mod}}$. But if we keep both $M,N$ variable, we get something that is not a functor, but which still has functorial properties. \begin{prop}[] Let $M \stackrel{u}{\to}M'$ and $N \stackrel{v}{\to}N'$ be linear maps. There exists a \emph{unique} linear map \begin{align*} M \otimes N \stackrel{u \otimes v}{\to} M' \otimes_R N' \end{align*} such that \begin{align*} (u \times v)(m \otimes n) = u(m) \otimes v(n) \end{align*} for all $(m,n) \in M \times N$. \end{prop} We prove this by applying the universal property of the tensor product. \begin{proof} The map \begin{align*} b: M \times N \to M' \otimes_R N', \quad (m,n) \mapsto u(m) \otimes v(n) \end{align*} is bilinear, as it is the composition of the bilinear map $\beta'$ with the linear map $(u,v)$ \begin{align*} M \times N \stackrel{(u,v)}{\to} M' \times N' \stackrel{\beta}{\to} M' \otimes N' \end{align*} by the universal property, there exists a unique linear map $f: M \otimes N \to M' \otimes N$ such that \begin{align*} f(m \otimes n) = b(m,n) = u(m) \otimes v(n) \end{align*} \end{proof} So now we can use the symbol $\otimes$ to denote the space $M \otimes_R N$, elements $m \otimes n$, and functions $u \otimes v$. \begin{lem}[] Let $M,N$ be $R$-modules. Then \begin{align*} \id_M \otimes \id_N = \id_{M \otimes N} \end{align*} and if $M \stackrel{u}{\to} M' \stackrel{u'}{\to}M''$ and $N \stackrel{v}{\to}N' \stackrel{v'}{\to}N''$ are linear, then \begin{align*} (u' \otimes v') \circ (u \otimes v) = (u' \circ u) \otimes (v' \circ v) \end{align*} \end{lem} The proof is elementary and follows from uniqueness. In particular, if $u,v$ are isomorphisms, then $u \otimes v$ is an isomorphism with inverse $u^{-1} \otimes v^{-1}$. We can thus say that the tensor product is a functor from the category $R_{\text{Mod}} \times R_{\text{Mod}} \to R_{\text{Mod}}$ that sends pairs of $R$-modules $(M,N)$ to $M \otimes N$ and a morphism $(u,v)$ to $u \otimes v$.
 We will start to compute the tensor product for simple cases. (Such as free modules like vector spaces). More generally, we will show how the tensor product behaves bilinearly with respect to direct sums. Given a family $(M_i)_{i \in I}$ of $R$-modules, recall that $\bigoplus_{i \in I}M_i$ is an $R$-module with linear maps \begin{align*} \phi_j: M_j \to \bigoplus_{i \in I}M_i, \quad \psi_j: \bigoplus_{i \in I}M_i \to M_j\\ x_j \mapsto (0,\ldots,0,x_j,0,\ldots,0), \quad (m_i)_{I \in I} \mapsto m_j \end{align*} which satisfy \begin{align*} \psi_j \circ \phi_j = \id_{M_j} \quad \text{and} \quad \bigoplus_{i \in I}(\phi_j \circ \psi_j) = \id_{\bigoplus_{i \in I}M_j} \end{align*} The direct sum consists of sums $\sum_{i \in I}'m_i$, where all but finitely many $m_i$ are zero. For finite index sets $I$, the direct sum is equal to the direct product. \begin{prop}[ Distributivity''] \label{prop:distributivity-tensor} Let $M,N$ be $R$-modules. \begin{enumerate} \item For any collection $(M_i)_{i \in I}$, (resp. $(N_i)_{i \in I}$) and any $j \in I$ there is an isomorphism of $R$-modules \begin{align*} &\bigoplus_{i \in I}(M_i \otimes N) \to \left(\bigoplus_{i \in I}M_i\right) \otimes N\\ &m_j \otimes n \mapsto \phi_j(m_j) \otimes n\\ \text{resp.} \quad &\bigoplus_{i \in I}(M \otimes N_i) \to M \otimes \left(\bigoplus_{i \in I}N_i\right)\\ &m \otimes n_j \mapsto m \otimes \phi_j(n_j) \end{align*} \item The map $M \otimes_R R \to M$ (resp. $R \otimes_R N \to N$) \begin{align*} m \otimes r \mapsto rm , \quad (\text{resp.} \quad r \times n \mapsto rn) \end{align*} is an isomorphism of $R$-modules. \end{enumerate} \end{prop} To make the proof a little shorter (and because this is useful on its own), we first prove the following \begin{lem}[] Let $(V_i)_{i \in I}$ be a collection of $R$-modules and $W$ another $R$-module. Then there is a bijection \begin{align*} \Hom_R\left( \bigoplus_{i \in I}V_i, W \right) \iso \bigoplus_{i \in I}\Hom_R(V_i,W) \end{align*} so to define a linear map from the direct sum is equivalent to defining a linear map from each component. \end{lem} \begin{proof} Set \begin{align*} &\Phi:\Hom_R\left( \bigoplus_{i \in I}V_i, W \right) \to \bigoplus_{i \in I}\Hom_R(V_i,W), \quad f \mapsto \bigoplus_{i \in I} (f \circ \phi_i)\\ &\Psi: \bigoplus_{i \in I}\Hom_R(V_i,W) \to \Hom_R\left( \bigoplus_{i \in I}V_i, W \right), \quad \bigoplus_{i \in I}f_i \mapsto \bigoplus_{i \in I}(f_i \circ \psi_j) \end{align*} and show that they are inverses of eachother. \end{proof} \begin{proof}[Proof Proposition] We begin with (b). Since the module structure is a bilinear map $R \times M \to M$, there exists a linear map $R \otimes M \stackrel{u}{\to}M$ with $u(r \otimes m) = rm$. Conversely, define the linear map. \begin{align*} v: M \to R \otimes M, \quad m \mapsto 1 \otimes m \end{align*} Because the pure tensors generate $R \otimes M$, it suffices to show that the composition $(v \circ u)$ is the identity on the pure tensors. \begin{align*} (u \circ v)(m) = u(1 \otimes m) = m \quad \text{and} \quad (v \circ u)(r \otimes m) = v(rm) = 1 \otimes rm = r(1 \otimes m) = r \otimes m \end{align*} which shows $M \iso R \otimes M$. Now for (a). first check the the map \begin{align*} m_j \otimes n \mapsto \phi_j(m_j) \otimes n \end{align*} is well-defined. We define \begin{align*} u_j = \phi_J \otimes \id_n \end{align*} and by the lamma, there is a map \begin{align*} f_j: M_j \otimes N \to \left( \bigoplus_{i \in I}M_i \right) \otimes N \end{align*} as required. To check that $f$ is an isomorphism, we define the bilinear map \begin{align*} \bigoplus_{i \in I}M_i \times N \to \bigoplus_{i \in I}(M_i \otimes N), \quad ((m_i)_{i \in I},n) \mapsto \sum_{i}' m_i \otimes n \end{align*} which is well-defined because all but finitely many $m_i$ are zero. By the universal propety, this defines a linear map \begin{align*} g: \left( \bigoplus_{i \in I}M_i \right) \otimes N \to \bigoplus_{i \in I}M_i \otimes N \end{align*} One easily checks that $f$ and $g$ are inverses of eachother: \begin{align*} (f \circ g)((m_i)_{i \in I} \otimes n) &= f\left( \sum_{i \in I}'m_i \otimes n \right) \\ &= \sum_{i\in I}' f_i(m_i \otimes n)\\ &= \sum_{i\in I}' \phi_i(m_i) \otimes n\\ &= \left( \sum_{i \in I}' \phi_i(m_i) \otimes n \right)\\ &= (m_i)_{i\in I} \otimes n \end{align*} and for the other way, it suffices to show this for generators of the module: \begin{align*} (g \circ f)(m_j \otimes n) = g(\phi_j(m_j) \otimes n) = m_j \otimes n, \quad \text{for} \quad j \in I \end{align*} \end{proof} \begin{cor}[] If $M$ is free with basis $(m_i)_{i \in I}$ and $N$ is free with basis $(n_j)_{j \in J}$, then $M \times N$ is free with basis $(m_i \times n_j)_{(i,j) \in I \times J}$. In particular, if $K$ is a field and $V,W$ finite-dimensional $K$-vector spaces. Then so is $V \times_K W$ and \begin{align*} \dim(V \otimes_K W) = (\dim V) (\dim W) \end{align*} \end{cor} \begin{proof}[] Let $M_i = R e_i, N_j = R f_j$ so that $M = \bigoplus_{i\in I}M_i$, $N = \bigoplus_{j \ni J}N_j$. By the proposition, we get an isomorphism \begin{align*} M \otimes N = \left( \bigoplus_{i\in I}M_i \right) \otimes \left( \bigoplus_{j \in J}N_j \right) \iso \bigoplus_{(i,j) \in I \times J} M_i \otimes N_j \end{align*} From the second part of the proposition, we see that \begin{align*} M_i \otimes N_j = Re_i \otimes N_j \iso N_j \end{align*} so it is free of rank $1$. Since it is generated by $e_i \otimes f_j$, this element is a basis, the above means that $M \otimes N$ is free with basis $(e_i \otimes f_j)_{(i,j) \in I \times J}$. \end{proof} \begin{xmp}[] Let $K$ be a field and take $M = N = K^{2}$ with basis $(e_1,e_2)$. Then $M \otimes_K N$ is a $K$-vector space of dimension $4$ with basis \begin{align*} e_1 \otimes e_1, \quad e_1 \otimes e_2, \quad e_2 \otimes e_1, \quad e_2 \otimes e_2 \end{align*} using this we can determine the basis of pure tensors. For an example of a pure tensor that is not a basis element, let \begin{align*} v_1 = x_1 e_1 + x_2 e_2 \in M, \quad v_2 = y_1 e_1 + y_2 e_2 \in N, \quad x_i,y_i \in K \end{align*} then by bilinearity, we find \begin{align*} v_1 \otimes v_2 = z_{11}e_1 \otimes e_1 + z_{12} e_1 \otimes e_2 + z_{21} e_2 \otimes e_1 + z_{22} e_2 \otimes e_2 \end{align*} where $z_{ij} = x_i y_i$. So a general element of the form \begin{align*} v = ae_1 \otimes e_1 + b e_1 \otimes e_2 + c e_2 \otimes e_1 + d e_2 \otimes e_2 \end{align*} is a pure tensor if there exist $x_1,x_2,y_1,y_2$ such that $a,b,c,d$ satisfy \begin{align*} a = x_1y_1, \quad b = x_1y_2, \quad c = x_2y_1, \quad d = x_2y_2 \end{align*} Which requires that $ad - bc = 0$. \end{xmp} Note for pure tensors, the representation $v = v_1 \otimes v_2$ is not unique, as we could also write $v = xv_1 \otimes x^{-1}v_2$ for a unit $x \in K^{\times}$.
 \begin{prop}[] Let $M,N$ be $R$-modules. There exists a unique isomorphism \begin{align*} M \otimes N \stackrel{u_{M,N}}{\to} N \times M \end{align*} such that $u_{M,N}(m \otimes n) = n \otimes m$ for all $(m,n) \in M \times N$. \end{prop} \begin{proof} As usual, we start by defining a bilinear map \begin{align*} M \times N \to N \otimes M, \quad (m,n) \mapsto n \otimes m \end{align*} so there exists a unique linear map \begin{align*} u_{M,N}: M \otimes N \to N \otimes N, \quad \text{such that} \quad u_{M,N}(m \otimes n) = n \otimes n \end{align*} The property $u_{M,N} \circ u_{N,M} = \id_{N \otimes M}$ is easy to check as it suffices to check it on the pure tensors (which are generators). \end{proof} We now consider \emph{associativity} of the tensor product: \begin{prop}[] Let $M,N,P$ be $R$-modules. There exists a unique isomorphism \begin{align*} (M \otimes N) \otimes P \stackrel{v}{\to} M \otimes (N \otimes P) \end{align*} such that \begin{align*} v((m \otimes n) \otimes p) = m \otimes (n \otimes p) \end{align*} for all $(m,n,p) \in M \times N \times P$. We therefore from now on write $M \otimes N \otimes P$ without any parenthesis to denote either construction. \end{prop} \begin{proof} The proof not as trivial as the previous ones. Set $T = M \otimes (N \otimes P)$. Since $\Hom((M \otimes N) \otimes P,T) \iso \Bil(M \otimes N,P;T)$, we just need to give a bilinear map. For any $p \in P$, consider the map \begin{align*} f_p: N \to N \otimes P, n \mapsto n \otimes p \end{align*} Let $g_p = \id_M \otimes f_p: M \otimes N \to T$ and let \begin{align*} b: (M \otimes N) \times P \to T \end{align*} this is bilinear and there fore the exists a unique linear map \begin{align*} v: (M \otimes N) \otimes P \to M \otimes (N \otimes P) \end{align*} such that \begin{align*} v((m \otimes n) \otimes p) &= b(m \otimes n,p)\\ &= g_p(m \otimes n)\\ &= m \otimes f_p(n) = m \otimes (n \otimes p) \end{align*} Similarly, we construct in the linear map on the opposite drection and show that they are reciprocal. \end{proof}
 We have seen distributivity of the tensor product with direct sums, which makes the computation much easier. Howoever, we we often have more complicated constructions that make use of submodules. Say we have a submodule $M' \subseteq M$ and we know what $M' \otimes N$ and $\faktor{M}{M'} \otimes N$ look like. What can we say about $M \otimes N$? \begin{dfn}[] Let $M',M,M''$ be $R$-modules with linear maps \begin{align*} M' \stackrel{f}{\to} M \stackrel{g}{\to}M'' \end{align*} If we $\image f \subseteq \kernel g$ (or equivalently $g \circ f = 0$), then we say that this is a \textbf{complex}. If $\image f = \kernel g$, then this is an \textbf{exact sequence}. We say that a sequence of $R$-modules \begin{align*} \ldots \to M_{n+1} \stackrel{f_{n+1}}{\to} M_{n} \stackrel{f_n}{\to} M_{n-1} \to \ldots, \quad n \in \Z \end{align*} is a \textbf{complex/exact}, if it is a complex/exact for each $n\in \Z$. \end{dfn} The language of exact sequences lets us express basic notions of algebra in a compact manner. \begin{xmp}[] Let $f: M \to N$ be an $R$-linear map. \begin{itemize} \item $f$ is injective, if and only if $0 \to M \stackrel{f}{\to}N$ is exact. \item $f$ is surjective, if and only if $M \stackrel{f}{\to}N \to 0$ is exact. \item $f$ is an isomorphism if and only if $0 \to M \stackrel{f}{\to} N \to 0$ is exact. \item If $M \stackrel{f}{\to}N \stackrel{g}{\to}P \to \stackrel{h}{\to} Q$ is exact, then $f$ is surjective if and only if $h$ is injective. \end{itemize} \end{xmp} \begin{dfn}[] A \textbf{short exact sequence} of $R$-modules is an exact sequence of the form \begin{align*} 0 \to M' \stackrel{f}{\to} M \stackrel{g}{\to} M'' \to 0 \end{align*} In this case, $M'\iso \image f$ and $M'' \iso \coker f = \faktor{M}{\image f} = \faktor{M}{\kernel g}$. We say that $M$ is an \textbf{extension} of $M''$ \textbf{by} $M'$. \end{dfn} \begin{xmp}[] The standard example of a short exact sequence is therefore \begin{align*} 0 \to N \stackrel{\iota}{\to} M \stackrel{\pi}{\to} \faktor{M}{N} \to 0 \end{align*} In fact, all short exact sequences can be identified to be of this type, but the general setting can be more flexible. For example, when $M = M' \oplus M''$, we have a short exact sequence \begin{align*} 0 \to M' \to &M \to M'' \to 0\\ x \mapsto &(x,0)\\ &(x,y) \mapsto y \end{align*} The converse however is not true, as the sequence \begin{align*} 0 \to 2\Z/4\Z \stackrel{\iota}{\to} \Z/4\Z \stackrel{(\cdot 2)}{\to} \Z/2\Z \to 0 \end{align*} is exact, but we do not have $\Z/4\Z \iso \Z/2\Z \oplus \Z/2\Z$ \end{xmp} So there are more complicated modules that are built'' from $M',M''$ by such a short exact sequence than the direct sum. We know that if $M = M' \oplus M''$, then by distributiity (Proposition \ref{prop:distributivity-tensor}) we have a short exact sequence \begin{align*} 0 \to M' \otimes N \to M \otimes N \to M'' \otimes N \to 0 \end{align*} What about the general case? \begin{prop}[Right-exactness of tensor product] If we have a short exact sequqnce \begin{align*} 0 \to M' \stackrel{f}{\to} M \stackrel{g}{\to} M'' \to 0 \end{align*} then then for any $R$-module $N$, there is an exact sequence \begin{align*} M' \otimes N \stackrel{f \otimes \id_N}{\to} M \otimes N \stackrel{g \otimes \id_N}{\to} M'' \otimes N \to 0 \end{align*} Note that $f \otimes \id_N$ might not always injective. \end{prop} We need to check that $\image f \otimes \id_N = \kernel g \otimes \id_N$ and that $g \otimes \id_n$ is surjective. The second part is easy. Because $M'' \otimes N$ is generated by elements $m'' \otimes n$ and $g$ is surjective, means that the generators are in $\image g \otimes \id_n$. The inclusion $\image f \otimes \id_N \subseteq \kernel g \otimes \id_n$ is also easy. Because $g \circ f = 0$, we have \begin{align*} (g \otimes \id_N) \circ (f \otimes \id_N) = (g \circ f) \otimes \id_N = 0 \end{align*} The converse inclusion is not obvious and we use a lemma for that. \begin{lem}[] A sequence of $R$-modules \begin{align*} N' \stackrel{f}{\to} N \stackrel{g}{\to} N'' \to 0 \end{align*} is exact if and only if, for any $R$-module $P$, the sequence \begin{align*} 0 \to \Hom(N'',P) \stackrel{h_P(g)}{\to} \to \Hom(N,P) \stackrel{h_P(f)}{\to} \Hom(N',P) \end{align*} is exact, where as usual $h_p(f) = - \circ f$. \end{lem} \begin{proof}[Proof Proposition] By the assumption in the proposition and the Lemma, for any $Q$, the following sequence is exact. \begin{align*} 0 \to \Hom(M'',Q) \stackrel{h_Q(g)}{\to} \Hom(M,Q) \stackrel{h_Q(f)}{\to} \Hom(M',Q) \end{align*} By setting $Q = \Hom(N,P)$ and using the $\Hom$-Tensor adjunction \begin{align*} 0 \to \Hom(M \otimes N, P) \iso \Hom(M,\Hom(N,P)), \quad f \mapsto (n \mapsto (m \mapsto f(m \otimes n))) \end{align*} (Similarly for $M',M''$), we get an exact sequence \begin{align*} 0 \to \Hom(M'' \otimes N,P) \stackrel{u}{\to} \Hom(M \otimes N,P) \stackrel{u}{\to} \Hom(M' \otimes N,P) \end{align*} if we can show that $u = h_P(g \otimes \id_N)$ and $v = h_P(f \otimes \id_N)$, then we can apply the Lemma in the opposite direction to \begin{align*} 0 \to \Hom(M'' \otimes N,P) \stackrel{h_P(g \otimes \id_N)}{\to} \Hom(M \otimes N,P) \stackrel{h_P(f \otimes \id_N)}{\to} \Hom(M' \otimes N,P) \end{align*} in order to obtain an exact sequence \begin{align*} M \otimes N \stackrel{f \otimes \id_N}{\to} M \otimes N \stackrel{g \otimes \id_N}{\to} M'' \otimes N \to 0 \end{align*} It remains to check the claim about $u$ and $v$.