[proofplan] We prove exactness directly from the [universal property of the tensor product](/theorems/3971) and the cokernel description of exactness. Surjectivity of $\operatorname{id}_M \otimes \beta$ follows by lifting the second tensor factor along the surjective map $\beta$. To identify the kernel, we quotient $M \otimes_R N$ by the image of $M \otimes_R N'$ and construct mutually inverse maps between this quotient and $M \otimes_R N''$. [/proofplan] [step:Introduce the induced maps and verify the composite is zero] Let \begin{align*} T' &:= M \otimes_R N', & T &:= M \otimes_R N, & T'' &:= M \otimes_R N''. \end{align*} Define the induced homomorphisms of abelian groups \begin{align*} a := \operatorname{id}_M \otimes \alpha &: T' \to T, \\ b := \operatorname{id}_M \otimes \beta &: T \to T''. \end{align*} For every $m \in M$ and $n' \in N'$, \begin{align*} (b \circ a)(m \otimes n') = b(m \otimes \alpha(n')) = m \otimes \beta(\alpha(n')) = m \otimes 0 = 0, \end{align*} because exactness of \begin{align*} N' \xrightarrow{\alpha} N \xrightarrow{\beta} N'' \end{align*} gives $\beta \circ \alpha = 0$. Since elementary tensors generate $T'$ as an abelian group, $b \circ a = 0$. Hence \begin{align*} \operatorname{im}(a) \subseteq \ker(b). \end{align*} [/step] [step:Prove that $\operatorname{id}_M \otimes \beta$ is surjective] Let $m \otimes n'' \in T''$ be an elementary tensor, with $m \in M$ and $n'' \in N''$. Since the sequence \begin{align*} N \xrightarrow{\beta} N'' \to 0 \end{align*} is exact, the map $\beta: N \to N''$ is surjective. Therefore there exists $n \in N$ such that $\beta(n) = n''$. Then \begin{align*} b(m \otimes n) = m \otimes \beta(n) = m \otimes n''. \end{align*} Since elementary tensors generate $T''$ as an abelian group, every element of $T''$ lies in $\operatorname{im}(b)$. Thus $b$ is surjective. [/step] [step:Construct the quotient that measures exactness at $M \otimes_R N$] Let $Q$ be the cokernel of $a$ in the category of abelian groups: \begin{align*} Q := T / \operatorname{im}(a). \end{align*} Let \begin{align*} q: T \to Q \end{align*} be the quotient homomorphism. Since $b \circ a = 0$, the image $\operatorname{im}(a)$ is contained in $\ker(b)$, so $b$ factors uniquely through $q$. Thus there exists a unique homomorphism of abelian groups \begin{align*} \bar{b}: Q \to T'' \end{align*} such that \begin{align*} \bar{b}(q(x)) = b(x) \end{align*} for every $x \in T$. In particular, for $m \in M$ and $n \in N$, \begin{align*} \bar{b}(q(m \otimes n)) = m \otimes \beta(n). \end{align*} [/step] [step:Define an inverse map from $M \otimes_R N''$ to the quotient] We define a map \begin{align*} c: M \times N'' \to Q \end{align*} as follows. For $m \in M$ and $n'' \in N''$, choose $n \in N$ with $\beta(n) = n''$, which is possible because $\beta$ is surjective, and set \begin{align*} c(m,n'') := q(m \otimes n). \end{align*} This definition is independent of the chosen lift. Indeed, if $n_1,n_2 \in N$ satisfy $\beta(n_1) = n''$ and $\beta(n_2) = n''$, then \begin{align*} \beta(n_1 - n_2) = 0. \end{align*} Thus $n_1 - n_2 \in \ker(\beta)$. Exactness gives $\ker(\beta)=\operatorname{im}(\alpha)$, so there exists $n' \in N'$ such that \begin{align*} n_1 - n_2 = \alpha(n'). \end{align*} Therefore \begin{align*} q(m \otimes n_1) - q(m \otimes n_2) = q(m \otimes (n_1-n_2)) = q(m \otimes \alpha(n')) = q(a(m \otimes n')) = 0. \end{align*} Hence $q(m \otimes n_1)=q(m \otimes n_2)$. The map $c$ is additive in each variable and $R$-balanced. For example, if $r \in R$, $m \in M$, and $n'' \in N''$, choose $n \in N$ with $\beta(n)=n''$. Since $\beta$ is $R$-linear, \begin{align*} \beta(rn)=r\beta(n)=rn'', \end{align*} so $rn$ is a lift of $rn''$. Hence \begin{align*} c(mr,n'') = q(mr \otimes n) = q(m \otimes rn) = c(m,rn''). \end{align*} The additivity checks are the same lifting computation, using additivity of the [tensor product](/page/Tensor%20Product) in each variable. By the universal property of $M \otimes_R N''$, the balanced map $c$ induces a unique homomorphism of abelian groups \begin{align*} \gamma: T'' \to Q \end{align*} such that \begin{align*} \gamma(m \otimes n'') = c(m,n'') \end{align*} for every $m \in M$ and $n'' \in N''$. [/step] [step:Show the two quotient maps are inverse isomorphisms] We prove that $\bar{b}$ and $\gamma$ are inverse homomorphisms. First let $m \in M$ and $n'' \in N''$. Choose $n \in N$ such that $\beta(n)=n''$. Then \begin{align*} (\bar{b} \circ \gamma)(m \otimes n'') = \bar{b}(q(m \otimes n)) = m \otimes \beta(n) = m \otimes n''. \end{align*} Since elementary tensors generate $T''$, this gives \begin{align*} \bar{b} \circ \gamma = \operatorname{id}_{T''}. \end{align*} Conversely, let $m \in M$ and $n \in N$. Then $n$ is a lift of $\beta(n) \in N''$, so by the definition of $\gamma$, \begin{align*} (\gamma \circ \bar{b})(q(m \otimes n)) = \gamma(m \otimes \beta(n)) = q(m \otimes n). \end{align*} The elements $q(m \otimes n)$ generate $Q$ as an abelian group because elementary tensors generate $T$ and $q$ is surjective. Therefore \begin{align*} \gamma \circ \bar{b} = \operatorname{id}_Q. \end{align*} Thus $\bar{b}: Q \to T''$ is an isomorphism. [/step] [step:Identify the kernel and conclude right exactness] Since $Q=T/\operatorname{im}(a)$ and $\bar{b}:Q\to T''$ is an isomorphism satisfying $\bar{b}\circ q=b$, the kernel of $b$ is precisely the kernel of the quotient map $q$. Therefore \begin{align*} \ker(b)=\ker(q)=\operatorname{im}(a). \end{align*} Together with the surjectivity of $b$, this proves that \begin{align*} M \otimes_R N' \xrightarrow{\operatorname{id}_M \otimes \alpha} M \otimes_R N \xrightarrow{\operatorname{id}_M \otimes \beta} M \otimes_R N'' \to 0 \end{align*} is exact. Since the argument applies to every exact sequence \begin{align*} N' \to N \to N'' \to 0 \end{align*} of left $R$-modules, the functor \begin{align*} M \otimes_R - : R\text{-}\mathrm{Mod} \to \mathrm{Ab} \end{align*} is right exact. [/step]