Let $V_1, V_2, W_1, W_2, U_1, U_2 \in$ K-Vect, $V_1 \xrightarrow{\;\; \alpha_1 \;\; }W_1 \xrightarrow{\;\; \beta_1 \;\; }U_1, V_2 \xrightarrow{\;\; \alpha_2 \;\; }W_2 \xrightarrow{\;\; \beta_2 \;\; }U_2$ K-linear.

I'm trying to check that $(\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2)=(\beta_1 \alpha_1)\otimes(\beta_2 \alpha_2)$ via the universal property of the tensorial product.

On one hand, we have $V_1 \times V_2 \xrightarrow{\;\; \otimes \;\; }V_1 \otimes V_2$ such that $(v_1,v_2) \longmapsto v_1 \otimes v_2$, bilinear. And we define $V_1 \times V_2 \xrightarrow{\;\; \alpha_1 \times \alpha_2 \;\; }W_1 \otimes W_2$ as $(\alpha_1 \times \alpha_2)(v_1,v_2)=\alpha_1(v_1) \otimes \alpha_2(v_2)$. I just proved that the morphism $(\alpha_1 \times \alpha_2)$ defined as before is bilinear. So, by the universal property, $\exists!\ V_1 \otimes V_2 \xrightarrow{\;\; \alpha_1 \otimes \alpha_2 \;\; }W_1 \otimes W_2$ linear such that $(\alpha_1 \times \alpha_2)=(\alpha_1 \otimes \alpha_2) \circ \otimes$.

Onthe other hand, we have $W_1 \times W_2 \xrightarrow{\;\; \otimes ' \;\; }W_1 \otimes W_2$ such that $(w_1,w_2) \longmapsto w_1 \otimes w_2$, bilinear. And we define $W_1 \times W_2 \xrightarrow{\;\; \beta_1 \times \beta_2 \;\; }U_1 \otimes U_2$ as $(\beta_1 \times \beta_2)(w_1,w_2)=\beta_1(w_1) \otimes \beta_2(w_2)$ (is bilinear). So, by the universal property, $\exists!\ W_1 \otimes W_2 \xrightarrow{\;\; \beta_1 \otimes \beta_2 \;\; }U_1 \otimes U_2$ linear.

Now, we take the composition $V_1 \times V_2 \xrightarrow{\;\; \alpha_1 \times \alpha_2 \;\; }W_1 \otimes W_2 \xrightarrow{\;\; \beta_1 \times \beta_2 \;\; }U_1 \otimes U_2$ such that $(\beta_1 \times \beta_2)(\alpha_1 \times \alpha_2)(v_1,v_2)=(\beta_1 \times \beta_2)(\alpha_1(v_1) \otimes \alpha_2(v_2))=\beta_1 \alpha_1(v_1) \otimes \beta_2 \alpha_2(v_2)$ which is bilinear.

So, we have the bilinear morphisms $\otimes$, $(\beta_1 \times \beta_2)(\alpha_1 \times \alpha_2)$ and by the universal property there exists a unique linear morphism $V_1 \otimes V_2 \xrightarrow{\;\; \beta_1 \alpha_1 \otimes \beta_2 \alpha_2 \;\; }U_1 \otimes U_2$ such that $(\beta_1 \times \beta_2)(\alpha_1 \times \alpha_2)=(\beta_1 \alpha_1 \otimes \beta_2 \alpha_2)\otimes$. But also the unique linear morphism $V_1 \otimes V_2 \xrightarrow{\;\; (\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2) \;\; }U_1 \otimes U_2$ satisfy the property, then $$(\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2)=(\beta_1 \alpha_1)\otimes(\beta_2 \alpha_2)$$

I was going to ask for a hint to prove this but as i typed this message i think i got the result. Is this correct? Am i missing something? Any suggestion or comments are welcome, thanks.

  • 1
    $\begingroup$ In your composition, you want $\beta_1 \otimes \beta_2$, not $\beta_1 \times \beta_2$. And in the last paragraph, you might want to explain why "But also the unique linear morphism $V_1 \otimes V_2 \xrightarrow{\;\; (\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2) \;\; }U_1 \otimes U_2$ satisfy the property" (and remove the "unique" here, as it is not to the point). Other than that, the proof is correct! $\endgroup$ – darij grinberg Sep 5 '13 at 4:33
  • $\begingroup$ I mean, as $(\beta_1 \otimes \beta_2)$ is unique by the univ. property and also $(\alpha_1 \otimes \alpha_2)$ is unique then the composition is unique and it satisfies that $(\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2)(v_1 \otimes v_2)=(\beta_1 \otimes \beta_2)(\alpha_1(v_1) \otimes \alpha_2(v_2))=\beta_1 \alpha_1(v_1) \otimes \beta_2 \alpha_2(v_2)$ but by the universal property the only one satisfing that is the morphism $\beta_1 \alpha_1 \otimes \beta_2 \alpha_2$, then they must be equal. Right, there's no way to compose $(\beta_1 \times \beta_2)(\alpha_1 \times \alpha_2)$ thanks. $\endgroup$ – Lix Sep 5 '13 at 5:10

No, $(\beta_1 \otimes \beta_2)(\alpha_1 \times \alpha_2)$ doesn't make sense.

Recall the general definition of the tensor product of linear maps. I won't repeat it here.

Then we compute, just using the definitions, for $v_i \in V_i$:

$((\beta_1 \alpha_1) \otimes (\beta_2 \alpha_2))(v_1 \otimes v_2) = (\beta_1 \alpha_1)(v_1) \otimes (\beta_2 \alpha_2)(v_2)$

$=\beta_1(\alpha_1(v_1)) \otimes \beta_2(\alpha_2(v_2)) = (\beta_1 \otimes \beta_2)(\alpha_1(v_1) \otimes \alpha_2(v_2))$

$=((\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2))(v_1 \otimes v_2).$

Thus, the two linear maps $V_1 \otimes V_2 \to U_1 \otimes U_2$ are equal when composed with the canonical bilinear map $V_1 \times V_2 \to V_1 \otimes V_2$, hence equal (by the universal property).

You can also write down the proof without elements. First, prove the corresponding statement for products. Now look at the diagram:

$\begin{array}{c} V_1 \times V_2 & \rightarrow & V_1 \otimes V_2 \\ \alpha_1 \times \alpha_2 \downarrow ~~~~~~~~~~~~~&& ~~~~~~~~~~~~~ \downarrow \alpha_1 \otimes \alpha_2 \\ W_1 \times W_2 & \rightarrow & W_1 \otimes W_2 \\ \beta_1 \times \beta_2 \downarrow ~~~~~~~~~~~~~&& ~~~~~~~~~~~~~ \downarrow \beta_1 \otimes \beta_2\\ U_1 \times U_2 & \rightarrow & U_1 \otimes U_2 \end{array}$

It commutes by definition. The vertical composition on the right is $(\beta_1 \otimes \beta_2) (\alpha_1 \otimes \alpha_2)$. On the other hand, the vertical composition on the left is $(\beta_1 \times \beta_2) (\alpha_1 \times \alpha_2) = (\beta_1 \alpha_1) \times (\beta_2 \alpha_2)$, hence the vertical composition on the right is also $(\beta_1 \alpha_1) \otimes (\beta_2 \alpha_2)$.

  • $\begingroup$ i'm kind of confused, i defined $(\alpha_1 \times \alpha_2)$ from $V_1 \times V_2$ to $W_1 \otimes W_2$, the same with $(\beta_1 \times \beta_2)$. Are you defining new functions in the diagram? $\endgroup$ – Lix Sep 6 '13 at 1:43

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