Abstract Nonsense

Triangle Completions

Point of Post: In this post we discuss when certain mapping triangles of modules may be completed.

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Motivation

The concept I’d like to discuss in this post is the more theoretical way of viewing mappings, in regards to “factoring through” (i.e. completing a mapping triangle). Said differently, given mappings in and out of a given module, say $f:A\to B$ and $g:A\to C$ when can we find a morphism $h:C\to B$ which makes the resulting triangular diagram commute? We shall see that this will be of great importance to us when we start getting into some of the more advanced module theoretic topics. It mostly makes proofs easier by being able to, given a diagram, just decide whether or not we can complete it.

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Triangle Completions

So, let’s get the basic set-up down. Suppose that we have three left $R$-modules $M,N,L$ and we have the following diagram

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$\begin{matrix} M & & \\ {_f}\big\uparrow & & \\ N & \underset{g}{-\!\!\!\twoheadrightarrow} & L\end{matrix}\quad\mathbf{(1)}$

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where $f\in\text{Hom}_R(M,N)$ and $g\in\text{Hom}_R(N,L)$ (where we use, as is customary, the double-headed arrow for epimorphism). We’d like to know when we can complete this diagram in the sense that we can find a map $h\in\text{Hom}_R(L,M)$ such that the resulting diagram

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$\begin{matrix} M & & \\ {_f}\big\uparrow& \nwarrow{^h} & \\ N & \underset{g}{-\!\!\!\twoheadrightarrow}& L\end{matrix}\quad\mathbf{(2)}$

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Note that evidently if this were true, then given $x\in \ker g$ we have that $f(x)=h(g(x))=0$ and so $x\in\ker f$. In other words, it’s clear that a necessary condition for the existence of such an $h$ is that $\ker g\subseteq\ker f$. Conversely, if $\ker g\subseteq \ker f$ then the ‘obvious’ way to define a map $h:L\to M$ works. Namely, we know how to get from $N$ to $M$, and every element of $L$ is associated to an element of $N$, namely $\ell=g(n)$ for some $n\in N$. So, in a perfect world we’d just define $h(\ell)=f(n)$. Of course, the problem with this is well-definedness. But, the cool thing is that if $g(n)=g(n')$ then $n-n'\in\ker g$ and so $n-n'\in ker f$ so that $f(n)=f(n')$. So, the map is well-defined and clearly satisfies $h\circ g=f$. Thus, it remains to check that $h$ is an $R$– morphism. To see this we merely note if $\ell=g(n)$ then $r\ell=g(rn)$ so that, by independence of choice, we have that $h(r\ell)=f(rn)=rf(n)=h(\ell)$. Similarly, if $\ell=g(n)$ and $k=g(n')$ then $\ell+k=g(n+n')$ and so by well-definedness $h(k+\ell)=f(n+n')=f(n)+f(n')=h(k)+h(\ell)$. Summing this all up gives:

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Theorem: If $R$ is a ring and $M,N,L$ are left $R$-modules. Then, the diagram $\mathbf{(1)}$ may be completed to a diagram $\mathbf{(2)}$ if and only if $\ker g\subseteq\ker f$.

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Of course, we know categorically that we should be able to replace the above theorem by a dual theorem by replacing all the arrows, switching inclusions, and switching $\ker$ to $\text{im}$. Indeed, suppose we tried to complete the following diagram

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$\begin{matrix} M & & \\ {_f}\big\downarrow & & \\ N & \leftarrowtail & L\end{matrix}\quad\mathbf{(3)}$

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(where we use $\leftarrowtail$ for monomorphism) to a diagram of the form

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$\begin{matrix} M & & \\ {_f}\big\downarrow & \searrow{^h} & \\ N & \leftarrowtail & L\end{matrix}\quad\mathbf{(4)}$

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As last time there is clearly a necessary condition, for if $f(x)\in\text{im }f$ then $f(x)=g(h(x))$  and so $f(x)\in\text{im }g$. Thus, we see that it’s necessary that$\text{im }f\subseteq\text{im }g$. Conversely, if this is true it’s clear that the ‘obvious’ way of defining a map $M\to L$ should work. Namely, given any $m\in M$ we know that $f(m)\in\text{im }f$ and since $\text{im }f\subseteq\text{im }g$ this implies that $f(m)=g(\ell)$ for some $\ell\in L$, and since $g$ is injective we know that this $\ell$ is unique. We can then define $h(m)=\ell$. It’s clear that this is well-defined. More formally what we have done is used the fact that since $g$ is injective it posseses a let inverse $g^{\leftarrow}:\text{im }g\to L$ and so what we are doing is defining $h:M\to L$ by $h=g^{\leftarrow}\circ f$ where we made vital use of $\text{im }f\subseteq\text{im }g$. Now, since evidently $g\circ h=g\circ g^{\leftarrow}\circ f=f$ it suffices to check that $h$ is a morphism. To see this it clearly suffices to prove that $g^{\leftarrow}$ is a $R$-morphism since then $h$ is the composition of $R$-morphisms. To do this we merely note that if $g(\ell)=n$ then $g(r\ell)=rn$ so that $g^{\leftarrow}(r n)=r\ell=rg^{\leftarrow}(n)$. Similarly, if $g(\ell)=n$ and $g(k)=n'$ then $g(\ell+k)=n+n'$ so that $g^{\leftarrow}(n+n')=\ell+k=g^{\leftarrow}(n)+g^{\leftarrow}(n')$.

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Summarizing:

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Theorem: Let $R$ be a ring and $M,N,L$ left $R$-modules. Then a diagram $\mathbf{(3)}$ may be extended to a diagram $\mathbf{(4)}$ i and only if $\text{im }f\subseteq\text{im }g$.

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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.