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If $0 \rightarrow X \xrightarrow{i} Y \xrightarrow{\pi} Z \rightarrow 0$ is a short exact sequence and $W$ is a vector space, then the sequence $$0 \rightarrow X \otimes W \xrightarrow{i\otimes \text{id}_W}Y \otimes W\xrightarrow{\pi \otimes \text{id}_W}Z \otimes W \rightarrow 0 $$ is also exact. In particular,if $X \subset Y$ is a subspace and $W$ is arbitrary, then show that $$\frac{Y \otimes W}{X \otimes W} \simeq \frac{Y}{X} \otimes W$$

Before proving it, here is a small lemma that we shall need.

If $\phi:W \to Y$ and $\psi: X \to Z$ are injective linear maps, then $$\phi \otimes \psi: W \otimes X \to Y \otimes Z$$ is also injective.

Proof: Suppose that $\{w_i\}_{i \in I}$ is a basis for $W$. Let $v \in W \otimes X$. Then there exists a unique $\{x_i\}_{i \in I} \subset X$ such that $v =\sum_{i \in I}w_i \otimes x_i$. It is easy to see that $\{\phi(w_i)\}_{i \in I}$ (All but finitely many terms are non zero) is a linearly independent set in $Y$. If $\phi \otimes \psi(v)=0$, then we have $\sum_{i=1}^n \phi(w_i) \otimes \psi(x_i)=0$. Let $y_i^* \in Y^*$ be such that $y_i^*(\phi(w_j))=\delta_{ij}$ for each $i=1,2,\ldots,n$ and $j=1,2,\ldots,n$. Let $z^* \in Z^*$ be arbitrary. Then $$0=y_j^* \otimes z^*\left(\sum_{i=1}^n \phi(w_i) \otimes \psi(x_i)\right)=z^*(\psi(x_j))$$

Since this holds for all such $z^*$, we must $\psi(x_j)=0$ for each $j=1,2,\ldots,n$ which further implies that $x_j=0$ for each $j$, as $\psi$ is one to one. Thus, $v=0$.

From the above lemma, we see that $i \otimes \text{id}_W$ is one-to-one. Since $\pi$ is surjective, $\pi \otimes \text{id}_W$ is also surjective. Now we just need to show the exactness in the middle. Now suppose that $\{w_i\}_{i \in I}$ is a basis for $W$. Let $v \in \text{Image}(i \otimes \text{id}_W)$.Then there exists unique $\{y_i\}_{i \in I} \subset Y$ such that $v=\sum_{i \in I} y_i \otimes w_i$. Also there exists $u \in X \otimes W$ such that $v=i \otimes \text{id}_W (u)$. Moreover there exists unique $\{x_i\}_{i \in I}$ such that $u=\sum_{i \in I}x_i \otimes w_i$. Thus, we have $$\sum_{i \in I}y_i\otimes w_i=i \otimes \text{id}_W\left(\sum_{i \in I}x_i \otimes w_i \right)=\sum_{i \in I}i(x_i) \otimes w_i$$ which gives that $$\sum_{i \in I}\left(y_i-i(x_i)\right) \otimes w_i=0$$ By a similar argument as we did above, since $\{w_i\}_{i \in I}$ are linearly independent, we have that $i(x_i)=y_i$ for each $i \in I$. Since $\text{Image}(i)=\text{Kernel}(\pi)$, we have $\pi(y_i)=0$ for each $i \in I$. Thus, $\pi \otimes \text{id}_W(v)=0$.

For the other direction suppose that $v \in \text{Kernel}(\pi \otimes \text{id}_W)$. Then $v$ can be written as $\sum_{i \in I}y_i \otimes w_i$ for a unique $\{y_i\}_{i \in I} \subset Y$. Hence we have $\left( \pi \otimes \text{id}_W \right)(v)=\sum_{i \in I}\pi(y_i) \otimes w_i=0$. Again, since $\{w_i\}_{i \in I}$ are linearly independent, $\pi(y_i)=0$. So, $y _i \in \text{Kernel}(\pi)=\text{Image}(i)$ and hence can be written as $i(x_i)$ for some $x_i \in X$. Thus, $$v=\sum_{i \in I}i(x_i) \otimes w_i=i \otimes \text{id}_W \left(\sum_{i \in I}x_i \otimes w_i\right)$$

For the second part: We show the isomorphism between $\frac{Y \times W}{X \otimes W}$ and $Z \otimes W$ by exhibiting a right inverse from $\frac{Y \times W}{X \otimes W}$ to $\frac{Y}{X} \otimes W$ and left inverse from $Z \otimes W$ to $\frac{Y \otimes W}{X \otimes W}$. How do I show this?

Thanks for the help!!

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Because your tensor product is exact, all you have to do is take the short exact sequences

$$0 \to A\otimes X \to B\otimes X \to (A\otimes X)/(B\otimes X)\to 0$$

$$0 \to A \otimes X \to B \otimes X \to B/A \otimes X \to 0 $$

to conclude. The first is the canonical SES for a quotient, the other is the tensored SES of the SES for $A$ and $B$.

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  • $\begingroup$ What about the way I am trying? How do I fix the issues that have ? $\endgroup$ – tattwamasi amrutam May 25 '17 at 2:19
  • $\begingroup$ You're having too much of an issue because you're looking at elements. Use universal properties. There is always a map from $(A\otimes X)/(B\otimes X)$ to $A/B\otimes X$ that completes the diagram of SES, and it is forcedfully an isomorphism in your casef. $\endgroup$ – Pedro Tamaroff May 25 '17 at 2:35

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