Basically the problem is asking to prove $$(M_1\coprod M_2)/(N_1\coprod N_2)\cong (M_1/N_1)\coprod (M_2/N_2)$$Where $M_i$ is an $R$-module and $N_i$ is a subdmodule of $M_i$. I did: an element in $(M_1\coprod M_2)/(N_1\coprod N_2)$ looks like $$\binom{m_1}{m_2}+\binom{N_1}{N_2}\mbox{ where } m_1\in M_1\mbox{ and }m_2\in M_2$$

But this is equal to: $$\left\{ \binom{m_1}{m_2}+\binom{n_1}{n_2}: n_1\in N_1\mbox{ and }n_2\in N_2 \right\}$$ and adding component-wise we get: $$\left\{\binom{m_1+n_1}{m_2+n_2}:n_1\in N_1 \mbox{ and } n_2\in N_2\right\}$$ Which is precisely an element in $(M_1/N_1)\coprod (M_2/N_2)$, so the sets are equal (and obviously isomorphic), so my question goes, what is wrong because I don't think they would ask me to prove "$\cong$" when in reality they are "$=$"


  • $\begingroup$ The "elements" of $(M_1\amalg M_2)/(N_1\amalg N_2)$ are sets of pairs as you describe. But the elements of $(M_1/N_1)\amalg(M_2/N_2)$ are not the sets you describe: they are pairs of sets: $$\binom{\{m_1+n_1\mid n_1\in N_1\}}{\{m_2+n_2\mid n_2\in N_2\}};$$so you are blurring the distinction by asserting that a set of pairs is the same thing as the pair of sets. $\endgroup$ – Arturo Magidin Mar 7 '12 at 20:54
  • $\begingroup$ This notation does not seem typical. $\endgroup$ – Dylan Moreland Mar 7 '12 at 22:30
  • $\begingroup$ @Dylan: The binary coproduct in the category of $R$-modules is just the direct product/direct sum. He seems to be writing the elements as "column vectors" instead of ordered pairs. Not typical, but easily understandable, I think. And of course, the elements of $M/N$ are the cosets $m+N$, which, "literally", are the sets $\{m+n\mid n\in N\}$. $\endgroup$ – Arturo Magidin Mar 7 '12 at 22:34

First: the objects in question are not literally equal: the elements of $$(M_1\coprod M_2)/(N_1\coprod N_2)$$ are congruence classes of elements of $M_1\coprod M_2$ modulo the submodule $N_1\coprod N_2$; in particular, they are sets of elements of $M_1\coprod M_2$; that is, sets of ordered pairs $(m_1,m_2)$, with $m_1\in M_1$ and $m_2\in M_2$.

By contrast, the elements of $$\frac{M_1}{N_1}\coprod \frac{M_2}{N_2}$$ are pairs of elements, one in $\frac{M_1}{N_1}$ and one in $\frac{M_2}{N_2}$; these are themselves sets of elements of $M_1$ and of elements of $M_2$, respectively.

So, the elements of $(M_1\coprod M_2)/(N_1\coprod N_2)$ are sets of pairs of the form $(m_1,m_2)$.

The elements of $(M_1/N_1)\coprod (M_2/N_2)$ are pairs $(S,T)$, where $S$ is a subset of $M_1$ and $T$ is a subset of $N_2$.

So one is a set of sets of pairs, the other is a set of pairs of sets. Different objects. You are blurring this distinction in your second and third steps.

So you don't have equality, but you may have isomorphism.

Second: the fact that you are using $\coprod$ instead of $\times$ or $\oplus$ suggests that you should focus instead on the properties of these objects rather than any specific construction.

Namely: the object $A\coprod B$ is an object, together with two morphisms $i\colon A\to A\coprod B$ and $j\colon B\to A\coprod B$, with the property that for every module $X$ and every pair of morphisms $f\colon A\to X$ and $g\colon B\to X$, there exists a unique morphism $F\colon A\coprod B\to X$ such that $f=i\circ F$ and $g=j\circ F$.

The object $A/N$ together with the morphism $\pi\colon A\to A/N$ is the object with $N\subseteq \mathrm{ker}(p)$ with the property that for every object $B$ and every morphism $f\colon A\to B$, if $N\subseteq \mathrm{ker}(f)$, then there exists a unique $\overline{f}\colon (A/N)\to B$ such that $f = \overline{f}\circ \pi$.

Now, let $M_1$ and $M_2$ be modules, and $N_1$, $N_2$ submodules of $M_1$ and $M_2$, respectively. We can map $M_1$ to $(M_1/N_1)$, and from $(M_1/N_1)$ to $(M_1/N_1)\coprod (M_2/N_2)$. Likewise, we can map $M_2$ to $(M_2/N_2)$ and from $(M_2/N_2)$ to $(M_1/N_1)\coprod (M_2/N_2)$.

This gives us a pair of maps, one, let's call it $\pi_1$ from $M_1$ to $(M_1/N_1)\coprod (M_2/N_2)$; and one, let's call it $\pi_2$, from $M_2$ to $(M_1/N_1)\coprod (M_2/N_2)$. By the universal property of the coproduct, we therefore have a unique morphism $\Pi\colon M_1\coprod M_2\to (M_1/N_1)\coprod (M_2/N_2)$, such that $\Pi\circ i = \pi_1$ and $\Pi\circ j = \pi_2$.

Note that $\Pi$ is onto: this follows because $(M_1/N_1)\coprod (M_2/N_2)$ is generated by the images of $M_1/N_1$ and of $M_2/N_2$ inside the coproduct. Since the map $M_1\to M_1/N_1$ and the map $M_2\to M_2/N_2$ are both onto, then $\langle \pi_1(M_1),\pi_2(M_2)\rangle$ contains the images of $(M_1/N_1)$ and of $(M_2/N_2)$, and therefore the image of $\Pi$ is all of $(M_1/N_1)\coprod(M_2/N_2)$.

By the First Isomorphism Theorem, we conclude that $$\frac{M_1\coprod M_2}{\mathrm{ker}(\Pi)} \cong \frac{M_1}{N_1}\coprod\frac{M_2}{N_2}.$$

The above is valid, for example, in the category of Groups (with the free product in stead of the coproduct); for the case of modules, we can explicitly describe the kernel:

What is $\mathrm{ker}(\Pi)$? It is exactly $N_1\coprod N_2$. Indeed, it certainly contains the image of $N_1$ and the image of $N_2$; and given any object $(m_1,m_2)\in M_1\coprod M_2$, we have that $\Pi(m_1,m_2) = (m_1+N_1,m_2+N_2)$, which is equal to the zero element of $(M_1/N_1)\coprod (M_2/N_2)$ if and only if $m_1+N_1 = N_1$ and $m_2+N_2=N_2$, if and only if $m_1\in N_1$ and $m_2\in N_2$, if and only if $(m_1,m_2)\in N_1\coprod N_2$.

P.S. The isomorphism is much more natural if you view the objects as products rather than coproducts: we can map $M_1\prod M_2$ to both $M_1/N_1$ and to $M_2/N_2$, so that induces a map $M_1\prod M_2\to (M_1/N_1)\prod (M_2/N_2)$; the map is onto, and the kernel is precisely $N_1\prod N_2$. This holds in Groups as well.

  • $\begingroup$ Perfect! Thank you so much. $\endgroup$ – Daniel Montealegre Mar 7 '12 at 23:04
  • $\begingroup$ A couple of things: There are some minor typos in the paragraph when you talk about $A\coprod B$. Now a couple of questions: Using what you explained you say that there is a morphism from $M_1$ to $(M_1/N_1)\coprod (M_2/N_2)$ is the first map (the one from $M_1$ to $(M_1/N_1)$ ) the canonical epimorphism and then the one from $(M_1/N_1)$ to $(M_1/N_1)\coprod (M_2/N_2)$ then inclusion map (i.e., $m_1\mapsto (m_1,0)$)? $\endgroup$ – Daniel Montealegre Mar 8 '12 at 1:43
  • $\begingroup$ Yes... except that should be $m_1 \mapsto (m_1+N_1,0)$. Similarly, there is a map from $M_2$ to the coproduct, given by $m_2\mapsto (0,m_2+N_2)$. But I didn't want to use the explicit mappings; instead, using the "canonical mappings": Map $M_1$ to $M_1/N_1$ "canonically" (the canonical epimorphism), and then map $M_1/N_1$ to $(M_1/N_1)\coprod (M_2/N_2)$ "canonically" (using the canonical map $i\colon (M_1/N_1) \to (M_1/N_1)\coprod (M_2/N_2)$). By using only the "natural structure maps", the discussion also works in other contexts (e.g., groups). $\endgroup$ – Arturo Magidin Mar 8 '12 at 4:03

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