# Abstract Nonsense

## Relationship Between Hom and Limits (Modules)(Pt. II)

Point of Post: This is a continuation of this post.

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Now that we have done the one case, the other case should follow pretty (in taste) similarly. Namely, the only real difference is that now $\text{Hom}_R(M,\bullet)$ is a covariant, and not contravariant, functor and so none of the arrows get reversed. That said, for the sake of logical completeness we’ll go through precisely what we mean to say.

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We start with a set-up similar to what we had before. Namely, we assume we are given some preordered set $\left(\mathcal{A},\leqslant\right)$, only now we assume we are given some inverse system $\left(\left\{N_\alpha\right\}_{\alpha\in\mathcal{A}},\left\{f_{\alpha,\beta}:N_\beta\to N_\alpha\right\}_{\alpha,\beta\in\mathcal{A},\;\alpha\leqslant\beta}\right)$ with inverse limit $\left(\varprojlim N_\alpha,\{p_\alpha\}\right)$ (where, as in the previous case, we assume we are given a general inverse limit and not necessarily the standard one, this makes things less messy). We then begin by noticing that each map $f_{\alpha,\beta}:N_\beta\to N_\alpha$ induces a map $f_{\alpha,\beta}^\ast:\text{Hom}_R(M,N_\beta)\to\text{Hom}_R(M,N_\alpha)$ given by postcomposition, or said differently,

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$\text{Hom}_R(M,N_\beta)\ni g\overset{f_{\alpha,\beta}^\ast}{\mapsto}f_{\alpha,\beta}\circ g\in\text{Hom}_R(M,N_\alpha)$

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Now, what we claim is that $\left(\left\{\text{Hom}_R(M,N_\alpha)\right\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}^\ast:\text{Hom}_R(M,N_\beta)\to\text{Hom}_R(M,N_\alpha)\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta}\right)$ is an inverse system over $\left(\mathcal{A},\leqslant\right)$. Of course, the same caveat goes for this case as the previous. Namely, while true regardless, it is important to note that in general this is nothing more than an inverse system of $\mathbb{Z}$-modules (e.g. abelian groups) but if we assume that $R$ is commutative it is an inverse system of $R$-modules. With this in mind, to check that this is an inverse system it suffices in both cases to merely check that $f_{\alpha,\alpha}^\ast=\text{id}_{\text{Hom}_R(M,N_\alpha)}$ and $f_{\alpha,\gamma}^\ast=f_{\alpha,\beta}^\ast\circ f_{\beta,\gamma}^\ast$ whenever $\alpha\leqslant\beta\leqslant\gamma$. The first of these is as simple as the previous case since $f_{\alpha,\alpha}^\ast(g)=f_{\alpha,\alpha}\circ g=\text{id}_{\text{Hom}_R(M,N_\alpha)}\circ g=g$ for all $g\in\text{Hom}_R(M,N_\alpha)$ so that $f_{\alpha,\alpha}^\ast$ is identity as desired. The other identity follows suit with the previous case except the noted non-switching of the composition order:

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$f_{\alpha,\beta}^\ast(f_{\beta,\gamma}^\ast(g))=f_{\alpha,\beta}^\ast(f_{\beta,\gamma}\circ g)=f_{\alpha,\beta}\circ f_{\beta,\gamma}\circ g=f_{\alpha,\gamma}\circ g=f_{\alpha,\gamma}(g)$

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Thus, ta-da, we now are sure that our system really is an inverse system where the maps are group homomorphisms (always) with the further nicety of being $R$-maps if $R$ is commutative. We are now in a position to state and prove the titular theorem:

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Theorem: Let $\left(\mathcal{A},\leqslant\right)$ be a preordered set and $\left(\{N_\alpha\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}:N_\beta\to N_\alpha\}_{\alpha,\beta\in\mathcal{A},\;\alpha\leqslant\beta}\right)$ some inverse system of left $R$-modules over $\left(\mathcal{A},\leqslant\right)$ with inverse limit $\left(\varprojlim N_\alpha,\{p_\alpha\}\right)$. Then, if $\left(\left\{\text{Hom}_R(M,N_\alpha)\right\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}^\ast:\text{Hom}_R(M,N_\beta)\to\text{Hom}_R(M,N_\alpha)\}_{\alpha,\beta\in\mathcal{A},\;\alpha\leqslant\beta}\right)$ is the inverse system of modules described above then $\left(\text{Hom}_R\left(M,\varprojlim N_\alpha\right),\{\pi_\alpha\}\right)$, with $\displaystyle \pi_\beta:\text{Hom}_R\left(M,\varprojlim N_\alpha\right)\to\text{Hom}_R\left(M,N_\beta\right)$ given by $\pi_\beta(g)=p_\beta\circ g$, is an inverse limit of this system. This is an inverse limit of abelian groups, and if $R$ happens to be commutative, then it is an inverse limit of $R$-modules.

Proof: Same as before, we let $A$ be either the ring $\mathbb{Z}$ or $R$ and go through the proof. We then begin by noting that the $\pi_\alpha$ do satisfy the desired compatibility with the $f_{\alpha,\beta}$ since

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$\pi_\beta(f_{\alpha,\beta}^\ast(g))=\pi_\beta(f_{\alpha,\beta}\circ g)=p_\beta\circ f_{\alpha,\beta}\circ g=p_\alpha\circ g=\pi_\alpha(g)$

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for any $g\in\text{Hom}_R\left(M,\varprojlim N\right)$ so that $\pi_\beta\circ f_{\alpha,\beta}^\ast=\pi_\alpha$ as desired. We now show that it has the unique lifting property. To this end we assume we are given a set of $A$-maps $g_\alpha:X\to\text{Hom}_R(M,N_\alpha)$ such that $f_{\alpha,\beta}^\ast\circ g_\beta=g_\alpha$. We then note that if we fix some $x\in X$ then we have a set of maps $g_\alpha(x):M\to N_\alpha$ such that $f_{\alpha,\beta}\circ g_\beta(x)=(f_{\alpha,\beta}^\ast\circ g_\beta)(x)=g_\alpha(x)$. Thus, by the definition of the inverse limit we are guaranteed a map $j_x:M\to\varprojlim N_\alpha$ such that $p_\alpha\circ j_x=g_\alpha(x)$. We then define a map $j:X\to\text{Hom}_R\left(M,\varprojlim N_\alpha\right)$ by $j:x\mapsto j_x$. To see that this is $A$-linear we merely note

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$p_\alpha\circ j_{ax+by}=g_\alpha(ax+by)=ag_\alpha(x)+bg_\alpha(y)=p_\alpha\circ(aj_x+bj_y)$

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for every $\alpha\in\mathcal{A}$, and so by definition of inverse limit we have that $j_{ax+by}=aj_x+bj_y$ as desired. Lastly, noting that $\pi_\alpha(j(x))=p_\alpha\circ j_x=g_\alpha(x)$ for all $x\in X$ and so $\pi_\alpha\circ j=g_\alpha$. Thus, $j$ is our desired map. To note that $j$ is unique we merely note that if $k$ were another such map, then $p_\alpha\circ j(x)=p_\alpha\circ k(x)$ for all $\alpha\in\mathcal{A}$, and thus $j(x)=k(x)$ for all $x\in X$, and so $j=k$. Thus, uniqueness follows. With this in all in mind we see we have proven the desired requirements. $\blacksquare$

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We note that if we consider the trivial direct or inverse system then the induced inverse system (in both cases) is trivial from where we recover our two previously proved relations between Hom, product, and coproduct. It’s important to note that in that post we gave counterexamples to the other “obvious” permutations of product and coproduct in each of the entries of Hom and on the outside of Hom. This of course shows that these possible relations with corproduct replaced by direct limit and product replaced by inverse limit are doomed in general as well.

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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.