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I am trying to construct a tensor product on $ \mathbb R^2$.

I have defined a bilinear map by $\phi:\mathbb R \times \mathbb R\to \mathbb R $, as $\phi(x,y) = xy$

Which clearly is a bilinear map. Now lets define the entire $X$-axis as the subspace. Hence the quotient space consists of all parallel lines to $X$ axis.

Now I am trying to find the bilinear map $f$ , which takes any pair of $(x, y)$ to its equivalence class - $f:\mathbb R \times \mathbb R \to \mathbb R \otimes \mathbb R$. But I am unable to do so, because $f(x, y)$ and $f(3x, y)$ must belong to same class, but by bilnearity they are not. Example, $f(x,y_0)=y_0$, but $f(3x, y_0) = 3f(x,y_0)=y_0$

Which can not be true as $f(x,y_0)=f(3x,y_0)$ as they both belong to same horizontal line (equivalence class).

So unless I find $f$, I can not get to $\bar \phi:\mathbb R \otimes \mathbb R\to \mathbb R$.

I just want to construct this function $\bar \phi$, so that I can see how $\bar \phi (f(x,y)) = \phi(x,y)$

Any pointer will be highly appreciated! I am stuck at finding $f$.

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1 Answer 1

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As $\Bbb R$-vector space, $\Bbb R$ has basis $\{1\}$, thus $\Bbb R\otimes\Bbb R$ has basis $1\otimes1$ with associated bilinear map $$ \phi:\Bbb R\times\Bbb R\longrightarrow\Bbb R\otimes\Bbb R, \qquad \phi(x,y)=xy(1\otimes1) $$

If $f:\Bbb R^2\rightarrow\Bbb R$ is the bilinear map $f(x,y)=xy$, then $\bar f:\Bbb R\otimes\Bbb R\rightarrow\Bbb R$, defined as $\bar f(z(1\otimes1))=z$ is linear and $\bar f\circ\phi=f$.

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