# Abstract Nonsense

## Homomorphism Groups of Products and Coproducts (Pt. I)

Point of Post: In this post we prove the following two common, and useful isomorphisms: $\displaystyle \text{Hom}_R\left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha,N\right)\cong\prod_{\alpha\in\mathcal{A}}\text{Hom}_R(M_\alpha,N)$ and $\displaystyle \text{Hom}_R\left(M,\prod_{\alpha\in\mathcal{A}}N_\alpha\right)\cong\prod_{\alpha\in\mathcal{A}}\text{Hom}_R(M,N_\alpha)$.

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Motivation

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We shall in the future have many occasions to deal with homomorphism groups (modules). In particular, we shall often deal with one of the Hom functors, and surprisingly often we shall have that in the free variable there is a product or coproduct. Consequently, it would be nice if we had some way of simplifying such Hom’s in terms of nicer groups. That is precisely the content of this post. It shall be good practice for us applying our notions of product and coproduct.

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Homomorphism Groups of Products/Coproducts

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Let $R$ be a ring and $\left\{M_\alpha\right\}$ a set of left $R$-modules. Then, for any left $R$-module $N$ we have the following:

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Theorem: $\displaystyle \text{Hom}_R\left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha,N\right)\cong\prod_{\alpha\in\mathcal{A}}\text{Hom}_R(M_\alpha,N)$ where the isomorphism is of $\mathbb{Z}$-modules if $R$ is non-commutative, and of $R$-modules then $R$ is commutative.

Proof: In what follows let $A$ denote either $\mathbb{Z}$ or $R$. This way we shall see that the proofs of when we are dealing with $\mathbb{Z}$-modules or $R$-modules are really the same.  We define the maps $\displaystyle p_{\alpha_0}:\text{Hom}_R\left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha,N\right)\to\text{Hom}(M_{\alpha_0},N)$ by $p_{\alpha_0}(f)=f\circ \iota_{\alpha_0}$ (where $\iota_\alpha$ is the natural inclusion)–it is easy to see that each $p_\alpha$ is an $A$-map.  We claim that with this set of morphisms, $\displaystyle \text{Hom}_R\left(\bigoplus_{\alpha\in\mathcal{A}},N\right)$ becomes a product of the left $A$-modules $\left\{\text{Hom}_R(M_{\alpha},N)\right\}$. Indeed, suppose that $g_\alpha:L\to \text{Hom}_R(M_\alpha,N)$ is a set of $A$-linear maps. With these maps in mind, define a map $\displaystyle g:L\to\text{Hom}_R\left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha,N\right)$ by the rule

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$\displaystyle (g(\ell))(x_\alpha))=\sum_{\alpha\in\mathcal{A}}(g_\alpha(\ell))(x_\alpha)$

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Evidently then $g$ is an $A$-map, and

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$((p_\alpha\circ g)(\ell))(x)=(p_\alpha(g(\ell)))(x)=(g(\ell))(\iota_\alpha(x))=(g_\alpha(\ell))(x)$

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But, this was true for arbitrary $x$ and so $(p_\alpha\circ g)(\ell)=g_\alpha(\ell)$ and since this was for arbitrary $\ell$ this tells us that $p_\alpha\circ g=g_\alpha$ as desired. Moreover, it’s clear that this $g$ is unique since it’s action on each $M_\alpha$ is determined, and so uniqueness follows from the definition of coproduct. Thus, we have proven that $\displaystyle \left(\text{Hom}_R\left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha,N\right),\{p_\alpha\}\right)$ is a product of $\left\{\text{Hom}_R(M_\alpha,N)\right\}$ from where the theorem follows from the uniqueness, up to isomorphism, of products. $\blacksquare$

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So, this theorem may seem a little strange at first, in two ways. Firstly, why does the coproduct become a product, and secondly can’t we find some way to take products out? The second one is particularly relevant since, as we shall see, there is a way to bring products out of the second entry, and moreover in the natural way, as a product.  Well, while it may not be a good intuitive reason why this is true, we can definitely justify why we should not have expected any of the other three permutations

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\displaystyle \begin{aligned}&\text{Hom}_R\left(\prod_{\alpha\in\mathcal{A}} M_\alpha,N\right)\cong \prod_{\alpha\in\mathcal{A}}\text{Hom}_R(M_\alpha,N)\\ &\text{Hom}_R\left(\bigoplus_{\alpha\in\mathcal{A}} M_\alpha,N\right)\cong \bigoplus_{\alpha\in\mathcal{A}}\text{Hom}_R(M_\alpha,N)\\ &\text{Hom}_R\left(\prod_{\alpha\in\mathcal{A}} M_\alpha,N\right)\cong\bigoplus_{\alpha\in\mathcal{A}}\text{Hom}_R(M_\alpha,N)\end{aligned}

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To be particular, it’s clear that any isomorphisms could be true. Why? Well, let in any case $\mathbb{Q}=R=M_\alpha=N$ and $\mathcal{A}=\mathbb{N}$. Then, (recalling that for unital rings $\text{Hom}_R(R,R)\cong R$) the first two would tell us that $\left(\mathbb{Q}^{\mathbb{N}}\right)^\ast\cong\mathbb{Q}^\mathbb{N}$ and $\left(\mathbb{Q}^{\oplus\mathbb{N}}\right)^\ast\cong\mathbb{Q}^{\oplus\mathbb{N}}$ and the last that $\left(\mathbb{Q}^{\mathbb{N}}\right)^\ast\cong\mathbb{Q}^{\oplus\mathbb{N}}$, and all of these contradict the fact that for infinite dimensional vector spaces the dual space has dimension strictly greater than the original space. Thus, in essence, to not contradict this fact we need the “operator” in the first entry to get “bigger” when it gets pulled out of the Hom, and clearly the only way this can happen is to go $\displaystyle \bigoplus\to\prod$. Of course, this also points the following corollary:

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Corollary: Let $R$ be a ring and $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ a set of left $R$-modules. Then,

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$\displaystyle \left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha\right)^\vee\cong\prod_{\alpha\in\mathcal{A}}M_\alpha^\vee$

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From the obvious equality $\text{Hom}_R(R,R)\cong R$ we are able to rederive the formula we proved in the dual modules post that $(R^n)^\vee\cong R^n$ for all finite $n$ since, of course, coproducts and products agree for finite index sets.

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print

November 12, 2011 -

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