# Abstract Nonsense

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

Point of Post: In this post we discuss how the Hom functor relates to limits. This shall serve, later on, to be a prime example of adjoint functors.

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Motivation

Now that we have properly defined notions of direct  and inverse limits of modules there are several natural questions we can ask, relating to previous topics. Perhaps one of the natural questions one might ask is how direct and inverse limits react with the Hom module (group). We should already have a good idea what’s going to happen based on the case when we are taking the direct and inverse limit over trivial directed and inverse systems. In other words, we already know what happens in the case of products and coproducts. This suggests that perhaps something of the form $\text{Hom}(\varinjlim \bullet,\bullet)\cong \varprojlim\text{Hom}(\bullet,\bullet)$ and $\text{Hom}(\bullet,\varprojlim\bullet)\cong\varprojlim\text{Hom}(\bullet,\bullet)$. Of course, we have to define what precisely what we mean here for, obviously as it stands, this makes absolutely no sense–what are the systems we are taking the limits over on the right hand side of both these isomorphisms? Regardless, the idea of why this “should” be true is clear enough. Namely direct limits are constructed to be such that mappings out them are completely determined by a set of mappings out of each of the individual terms of the limit, and similarly inverse limits are such that mappings into them are dtermined by a set of mappings into each of the individual factors. This roughly tells us that $\text{Hom}(\varinjlim\bullet,\bullet)\sim\text{Hom}(\bullet,\bullet)$ where $\sim$ denotes “related by some operation”, and similarly for the other entry with inverse limit. What suggests that in both cases we would get inverse limits is that there are natural projections maps $\text{Hom}(\varinjlim\bullet,\bullet)\to\text{Hom}(\bullet,\bullet)$ and $\text{Hom}(\bullet,\varprojlim\bullet)\to\text{Hom}(\bullet,\bullet)$ which heavily suggests we should be doing some kind of inverse limit (whenever one see mappings into things, inverse limits should be an immediate thought). Perhaps a bigger hint, or those who are in the know (or have done things in a slightly different order than I am posting), is that we know the maps $M_\alpha\to M_\beta$ and $M_\beta\to M_\alpha$ (where $\alpha\leqslant\beta$) in the $\varinjlim$ and $\varprojlim$ case respectively naturally induce maps $\text{Hom}_R(M_\beta,N)\to\text{Hom}_R(M_\alpha,N)$ and $\text{Hom}_R(M_\beta,N)\to\text{Hom}_R(M_\alpha,N)$ respectively since, as we’ve already discussed, $\text{Hom}_R(\bullet,N)$ and $\text{Hom}_R(N,\bullet)$ are contravariant and covariant functors respectively. This heavily suggests that we are going to be doing some kind inverse limit. Making all of this rigorous is slightly annoying, but this the basic idea. Either way, the proof shouldn’t seem to foreign since we have already tackled the special case of products and coproducts.

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Relationship Beween Hom and Limits

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Now that we have discussed what we plan on doing, let’s get right to it. We’ll start with the identity $\text{Hom}_R\left(\varinjlim M_\alpha,N\right)\cong \varprojlim\text{Hom}_R(M_\alpha,N)$. As was previously stated we have to first define what inverse system we are taking a limit over–to this end we suppose we have already been given our preordered set $\left(\mathcal{A},\leqslant\right)$ and our direct system $\left(\{M_\alpha\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}:M_\alpha\to M_\beta\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta}\right)$ and assume that $\left(\varinjlim M_\alpha,\{\varphi_\alpha\}\right)$ is our direct limit (note that we are assuming a direct limit of any form, not necessarily the standard, this makes things more clear). The key hint was (as was noted in the motivation) the fact that $\text{Hom}_R(\bullet,N)$ is naturally a contravariant functor. Indeed, each map $f_{\alpha,\beta}:M_\alpha\to M_\beta$ naturally induces a map $f_{\alpha,\beta}^\ast:\text{Hom}_R(M_\beta,N)\to\text{Hom}_R(M_\alpha,N)$ given by

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

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What we claim then is that $\left(\left\{\text{Hom}_R(M)_\alpha,N)\right\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}^\ast:\text{Hom}_R(M_\beta,N)\to\text{Hom}_R(M_\alpha,N)\}_{\alpha,\beta\in\mathcal{A},\;\alpha\leqslant\beta}\right)$ is an inverse system. The one small point I’ve glossed over is: an inverse system of what? Namely, we know that the Hom groups are, in general, just abelian groups (at least naturally) and aren’t $R$-modules unless we assume that $R$ is commutative. Of course, this distinction is unimportant since what we have created is (always) an inverse system of abelian groups (i.e. $\mathbb{Z}$-modules) and when permitting (i.e. when $R$ is commutative) an inverse system of $R$-modules. Moreover, the projection maps $\text{Hom}_R\left(\varinjlim M_\alpha,N\right)\to\text{Hom}_R(M_\alpha,N)$ will (always) be group homomorphisms, and when permitting will be $R$-maps. That said, while the distinction (insofar as the proofs are concerned) is artificial, it’s something important to keep in mind. Anyways, to verify that the system we have in mind is, in fact, an inverse system we must really only check that $f_{\alpha,\alpha}^\ast=\text{id}_{\text{Hom}_R(M_\alpha,N)}$ and $f_{\alpha,\beta}^\ast\circ f_{\beta,\gamma}^\ast=f_{\alpha,\gamma}^\ast$ whenenver $\alpha\leqslant\beta\leqslant\gamma$.The first of these is trivial since for any $g\in\text{Hom}_R(M_\alpha,N)$ one has that $f_{\alpha,\alpha}^\ast(g)=g\circ f_{\alpha,\alpha}=g\circ \text{id}_{M_\alpha}=g$, and so $f_{\alpha,\alpha}^\ast$ is identity as desired. Thus, it remains to show that composition works out the way we want it to.  But, this is just a simple computation, namely given any $g\in\text{Hom}_R(M_\gamma,N)$ we have that

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

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from where the desired conclusion follows.  So, now that we know that this is a direct system, what we claim is that $\text{Hom}_R\left(\varinjlim M_\alpha,N\right)$ is an inverse limit of this system, with the projections $\pi_\beta:\text{Hom}_R\left(\varinjlim M_\alpha,N\right)\to\text{Hom}_R(M_\beta,N)$ given by $\pi_\beta(g)=g\circ\varphi_\beta$. Indeed:

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Theorem: Let $\left(\mathcal{A},\leqslant\right)$ be preordered set and $\left(\{M_\alpha\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}:M_\alpha\to M_\beta\}_{\alpha,\beta\in\mathcal{A},\;\alpha\leqslant\beta}\right)$ be any direct system of $R$-modules with (some) direct limit $\left(\varinjlim M_\alpha,\{\varphi_\alpha\}\right)$. Then, $\left(\text{Hom}_R\left(\varinjlim M_\alpha,N\right),\{\pi_\alpha\}\right)$ is an inverse limit of the inverse system $\left(\left\{\text{Hom}_R\left(M_\alpha,N\right)\right\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}^\ast:M_\beta\to M_\alpha\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta}\right)$ (as described above) where this isomorphism is of abelian groups, and moreover of $R$-modules if $R$ is commutative.

Proof: Throughout the proof we let $A$ be the ring $\mathbb{Z}$ or $R$, so that we can do both cases at once. We begin by noting that the “projection” maps are $A$-maps that satisfy the desired equations since

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$f_{\alpha,\beta}(\pi_\beta(g))=f_{\alpha,\beta}(g\circ \varphi_\beta)=g\circ \varphi_\beta\circ f_{\alpha,\beta}=g\circ \varphi_\alpha=\pi_\alpha(g)$

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for all $\alpha\leqslant\beta$ and $g\in\text{Hom}_R\left(\varinjlim M_\alpha,N\right)$. Suppose now that we are given an $A$-module $X$ and a set of $A$-maps $g_\alpha:X\to\text{Hom}_R(M_\alpha,N)$ such that $f_{\alpha,\beta}^\ast\circ g_\beta=g_\alpha$. We then note that if we fix $x\in X$ the set $g_\alpha(x):M_\alpha\to N$ is a set of maps such that $f_\beta(x)\circ f_{\alpha,\beta}=(f_{\alpha,\beta}^\ast\circ g)(x)=g_\alpha(x)$. Thus, by the definition of the direct limit we are guaranteed a map $j_x:\varinjlim M_\alpha\to N$ such that $j_x\circ\varphi_\alpha=g_\alpha(x)$. Let $j:X\to\text{Hom}_R\left(\varinjlim M_\alpha,N\right)$ by defined by $j:x\mapsto j_x$. We next note that $j$ is an $A$-map since

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

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and, as we can recall, maps are completely determined by their precomposition with each $\varphi_\alpha$. Next, note that $\pi_\alpha\circ j(x)=g_\alpha(x)$ (by definition of $j(x)=j_x$) for each $x\in X$ so that $\pi_\alpha\circ j=g_\alpha$ for each $\alpha\in\mathcal{A}$. Thus, $j:X\to\text{Hom}_R\left(\varinjlim M_\alpha,N\right)$ is the desired map. Moreover, it’s clear that such a map is unique, for if $k:X\to\text{Hom}_R\left(\varinjlim M_\alpha,N\right)$ is another $A$-map such that $\pi_\alpha\circ k=g_\alpha=\pi_\alpha\circ j$ then we see for each $x\in X$ that $k(x)\circ\varphi_\alpha=(\pi_\alpha\circ k)(x)=(\pi_\alpha\circ j)(x)=j(x)\circ \varphi_\alpha$ and so the $j(x),k(x)$ agree on each $\varphi_\alpha$ and so, by the definition of the direct limit, must agree on $\varinjlim M_\alpha$. Thus, $j(x)=k(x)$, and since $x\in X$ was arbitrary we may conclude that $j=k$.  Thus, we see that each set of $A$-maps into the $\text{Hom}_R(M_\alpha,N)$ lifts uniquely to a map into $\text{Hom}_R\left(\varinjlim M_\alpha,N\right)$ which is compatible with the $\pi_\alpha$‘s. From this we may conclude that $\left(\text{Hom}_R\left(\varinjlim M_\alpha,N\right),\{\pi_\alpha\}_{\alpha\in\mathcal{A}}\right)$ is an inverse limit as desired. $\blacksquare$

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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.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.