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I am not sure this is standard terminology but the notes I am using has a mapping defined as follows. In particular I need help with how to write out the definition in some related exercises.

Let $R$ be a commutative ring with identity. Let $M,N,P$ be $R$ modules and let $\theta : Hom_A (M,N) \otimes_R P \rightarrow Hom_A (M,N \otimes_R P)$ be the canonical mapping.

Is the canonical mapping theta given by $(f,y) \rightarrow (x \rightarrow (x \rightarrow f(x \otimes y))$ for $f \in Hom_A (M,N)$ and $ y \in P$?

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The rule you have given does not make sense for a map $Hom_A (M,N) \otimes_R P \rightarrow Hom_A (M,N \otimes_R P)$ –  Mariano Suárez-Alvarez Sep 30 '11 at 17:28
    
Thanks for your response. Is there some well defined map that is similar to the one above that you are aware of could be a possible candidate in this situation? –  user7980 Sep 30 '11 at 17:35
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Dear user7980: I think there is a typo. Please check the display. Also you should try to define a linear map from a tensor product by a certain bilinear map. –  Pierre-Yves Gaillard Sep 30 '11 at 17:39
    
Thanks for the help. I think the problem was the map was missing a parenthesis at $f(x \otimes y)$. –  user7980 Sep 30 '11 at 17:45
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I think you mean the linear map induced by the bilinear map $(f,y) \mapsto (x \mapsto f(x) \otimes y)$. (See Mariano's answer.) –  Pierre-Yves Gaillard Sep 30 '11 at 17:49
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There is a map $$\theta : Hom_A (M,N) \otimes_R P \rightarrow Hom_A (M,N \otimes_R P)$$ such that $$\theta(f\otimes p)(m)=f(m)\otimes p$$ for all $f\in Hom_A(M,N)$, all $p\in P$ and all $m\in M$. This is the map one usually refers to as the canonical map in this context.

It makes for a good exercise to check that there is indeed such a map, by the way!

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