Exercise A0.III.6.12

Preliminaries

Lemma (L3.3)

Injective homomorphisms are monomorphisms in $\RMod.$

Proof

Indeed, they are injective as set functions and hence have left inverses (as set functions). $\blacksquare$

Lemma (L3.4)

Surjective homomorphisms are epimorphisms in $\RMod.$

Proof

Similarly, they are surjective as set functions and hence equality can be established in $\mathbf{Set},$ where right inverses exist for all surjections (assuming axiom of choice). $\blacksquare$

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Kernels

Proposition (K1)

Kernels exist in $R\mathrm{-Mod}.$

Proof

This proof is more abstract than necessary.

Let $\varphi: M \to N$ be a homomorphism and let $\alpha : P \to M$ be another homomorphism such that $\varphi \circ \alpha = 0.$ In other words $\varphi(\im \alpha) = 0.$

Consider the set

$$K = \{ u \in M \mid \varphi(u) = 0 \}.$$

Note that $\varphi(u) = \varphi(v) = 0$ implies $\varphi(u) + \varphi(v) = \varphi(u + v) = 0,$ and similarly $\varphi(ru) = r\varphi(u) = 0.$ So $K$ is a submodule of $M,$ hence let $i : K \to M$ be an inclusion of $K$ into $M.$

Let’s find $\overline{\alpha}: P \to K.$ Using the canonical decomposition we can factor $\alpha$ as a surjection $\alpha' : P \to \im \alpha$ followed by the inclusion $j : \im \alpha \to M.$ But since $\im \alpha$ itself is a submodule of $K$ (it is a module and, by hypothesis, it is a subset of $K$) we have yet another inclusion $\iota : \im \alpha \to K.$ Let $\overline{\alpha} = \iota \circ \alpha'.$

Now, the upper triangle of this diagram commutes by canonical decomposition $\alpha = j \circ \alpha'.$ The right triangle commutes since $\im \alpha \subseteq K \subseteq M.$ The left triangle commutes since we’ve defined $\overline{\alpha} = \iota \circ \alpha'.$ Thus, the whole diagram commutes. Hence, $\overline{\alpha}$ exists and fits the requirement that $i \circ \overline{\alpha} = \alpha.$

By L3.3 we know that $i$ is a monomorphism and hence for any other morphism $\beta : P \to K$ satisfying $i \circ \beta = \alpha$ we know that $\beta = \overline{\alpha}.$ So $\overline{\alpha}$ is unique.

Thus, $K$ satisfies the universal property of $\ker \varphi$ in $\RMod.$

$\blacksquare$

Proposition (K2)

Let $\varphi : M \to N$ be an $R$-module homomorphism. Then $\ker \varphi$ is trivial if and only if $\varphi$ is injective as a set function.

Proof

Suppose $\ker \varphi$ is trivial and let $a, b \in M$ such that $\varphi(a) = \varphi(b).$ Now, observe that

$$0 = \varphi(a) - \varphi(b) = \varphi(a-b),$$

and so $a-b \in \ker \varphi$ by construction of kernel in K1. But $\ker \varphi$ is trivial and hence $a - b = 0$ or $a = b,$ forcing $\varphi$ to be injective.

Conversely, suppose $\varphi$ is injective as a set function. Let $a \in M$ be such that $\varphi(a) = 0$ (that is, $a \in \ker \varphi$). Since $\varphi$ is a homomorphism, we know that $\varphi(0) = 0$. Since $\varphi$ is injective, $\varphi(0) = \varphi(a)$ implies $0 = a.$ So $\ker \varphi$ is trivial.

$\blacksquare$

Proposition (K3)

Monomorphisms in $\RMod$ are injective as set functions, and every injective morphism is a monomorphism.

Proof

From L3.3 we know that injective morphisms are monomorphisms.

Suppose $\varphi : M \to N$ is a monomorphism. Now, take the inclusion $i : \ker \varphi \to M$ and note that $\varphi \circ i = 0.$ But also, note that $0 \circ i = 0.$

Since $\varphi$ is a monomorphism, $i = 0$ and since inclusions are injective, $\ker \varphi$ is trivial. By K2 we know that this forces $\varphi$ to be injective.

$\blacksquare$

Proposition (K4)

Every monomorphism in $\RMod$ identifies its domain with the kernel of some morphism.

Proof

Let $\varphi: K \to M$ be a monomorphism. Since $\im \varphi$ is a submodule of $M$, we can form the quotient $M/\im \varphi$ with the projection $\pi : M \to M / \im \varphi.$ Note that $\ker \pi = \im \varphi.$

By canonical decomposition we know that the restriction $\varphi' : K \to \im \varphi$ is surjective. By K3 we know that it must be injective. Being a surjective and injective homomorphism it is a bijection. Hence, it is an isomorphism. So $K \cong \im \varphi = \ker \pi$ as required.

$\blacksquare$

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Cokernels

Proposition (C1)

Cokernels exist in $\RMod.$

Proof

Suppose $\varphi : M \to N$ is a homomorphism and let $\beta : N \to P$ be an arbitrary morphism such that $\beta \circ \phi = 0.$ Since $\im \varphi$ is a submodule of $N,$ we can form a quotient $N/\im \varphi$ with a surjection $\pi : N \to N/\im \varphi.$

Recall that by the universal property of quotients, the morphism $\beta : N \to P$ factors uniquely as $\overline{\beta} \circ \pi = \beta,$ as long as $\im \varphi \subseteq \ker \beta.$

Since composing $\beta \circ \phi$ gives the zero morphism, $\beta$ must send $\im \varphi$ to $0,$ so $\beta(\im \varphi) = 0,$ but this is equivalent to $\im \varphi \subseteq \ker \beta.$

What if there is some other morphism $\sigma : N/\im \varphi \to P$ be such that $\sigma \circ \pi = \beta?$ Well, since $\pi$ is an epimorphism (L3.4) it must be that $\sigma = \overline{\beta}.$

Thus, $N/\im \varphi$ satisfies the universal property of $\coker \varphi$ in $\RMod.$

$\blacksquare$

Proposition (C2)

Let $\varphi : M \to N$ be an $R$-module homomorphism. Then $\coker \varphi$ is trivial if and only if $\varphi$ is surjective as a set function.

Proof

Suppose $\coker \varphi = N/\im \varphi$ is trivial. The quotient map $\pi : N \to N/\im \varphi$ has $\im \varphi$ as its kernel, but since the codomain is trivial, its kernel must be the whole $N.$ So $N = \im \varphi$, so $\varphi$ is surjective.

Conversely, suppose $\varphi$ is surjective as a set function. So its image is the whole $N$, and $\coker \varphi = N/N$ is trivial.

$\blacksquare$

Proposition (C3)

Epimorphisms in $\RMod$ are surjective as set functions, and every surjective morphism is an epimorphism.

Proof

From L3.4 we know that surjective morphisms are epimorphisms.

Suppose $\varphi : M \to N$ is an epimorphism. Now, take the projection $\pi : N \to \coker \varphi$ and note that $\pi \circ \varphi = 0.$ But also, note that $0 \circ \varphi = 0.$

Since $\varphi$ is an epimorphism, $\pi = 0$ and since quotient projections are surjective, $\coker \varphi$ is trivial. By C2 we know that this forces $\varphi$ to be surjective.

Proposition (C4)

Every epimorphism in $\RMod$ identifies its codomain with the cokernel of some morphism.

Proof

Let $\varphi: N \to C$ be an epimorphism.

Consider $\ker \varphi$ with the inclusion $i : \ker \varphi \to N.$ We have the short exact sequence

$$0 \to \ker \varphi \overset{i}{\to} N \overset{\varphi}{\to} C \to 0,$$

and so $C = N/\ker \varphi = \coker i.$

$\blacksquare$