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I'm trying to understand more explicitly what the natural homomorphism is for an $R$-module $M$ to the dual of it's dual $(M^\ast)^\ast$.

It makes sense to compose the natural homomorphism twice by $$ M\stackrel{\phi}{\to} M^*\stackrel{\psi}{\to} (M^*)^* $$ and then just say the natural homomorphism is $\psi\circ\phi$. But what explicitly is $\phi$? To each $m\in M$, to what $f\in H^*$ do we associate it with?

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up vote 2 down vote accepted

There is no natural homomorphism $M\to M^*$, for exactly the same reasons as for vector spaces.

There is a natural map $\phi:M\to(M^*)^*$, though, of exactly the same form as that of vector spaces. It maps an element $m\in M$ to the linear function $\phi(m):M^*\to R$ such that $\phi(m)(f)=f(m)$ for all $f\in M^*$.

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Thanks!, I see where I went wrong. – Duine Mar 11 '12 at 4:09

There is no natural homomorphism $M \to M^*$.

Each element of $M^*$ is a function from $M$ to $R$. It's convenient to think of evaluation as just being a product: for $f \in M^*$ and $m \in M$, you have $fm \in R$.

Each element of $M^{**}$ is a function from $M^*$ to $R$. It's again convenient to think of evaluation as a product: for $\omega \in M^{**}$ and $f \in M^*$, $\omega f \in R$. If you want to map $M$ to $M^{**}$, can you think of any way to multiply an element $m \in M$ with an element $f \in M^*$ to get an element of $R$? ....

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