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Definition A.1.1

[definition] A category $\mathcal{C}$ consists of the following:

• a class $\operatorname{obj}(\mathcal{C})$ of objects,
• a set $\Hom_{\mathcal C}(A, B)$ of morphisms for every ordered pair $(A, B)$ of objects,
• an identity morphism $\id_A \in \Hom_{\mathcal C}{A, A}$ for each object $A$,
• a composition function $\Hom_{\mathcal C}(A, B) \times \Hom_{\mathcal C}(B, C) \to \Hom_{\mathcal C}(A, C)$ for every ordered triple $(A, B, C)$ of objects.

We also assume that morphisms are subjected to associativity. [/definition]

Paradigm A.1.2

[greenbox=Paradigm] The fundamental category to keep in mind is $\textbf{Sets}$ of sets. By Russell’s paradox, this category is not small yet is very concrete and familiar. [/greenbox]

Examples A.1.3

[example] The category $\textbf{Ab}$ of abelian groups (and group homomorphisms).

The category $\textbf{Groups}$ of groups (and group homomorphisms).

The category $\textbf{Rings}$ of rings (and ring homomorphisms).

The category $R{-}\textbf{mod}$ (resp. $\textbf{mod}{-}R$) of left (resp. right) $R$-modules.

A discrete category is a is ohe in which every morphism is an identity morphism. [/example]

Small categories A.1.4

[definition] A category $\mathcal C$ is small if $\obj(\mathcal C)$ is a set. [/definition]

[example] $\textbf{Sets}$ is not small by Russell’s paradox.

Similar argumentation shows the same for $\textbf{Ab}$. Given any set $X$, there is an obvious trivial abelian group on the singleton ${X}$.

Finally, since $\ZZ{-}\textbf{mod}$ has the same objects as $\textbf{Ab}$, we get that the category of $R$-modules in general is not small. [/example]

[definition] A monoid is a set $M$ equipped with an associative law of composition and an identity element. [/definition]

Note that a category with exactly one object $*$ is the same thing as a monoid (by making $M = \Hom(*, *)$). This is intended to mirror the relationship between groupoids and groups.

[definition] A morphism $f:B \to C$ is an isomorphism in $\mathcal C$ if there is a morphism $g:C \to B$ so that $gf = \id_B$ and $fg = \id_C$. [/definition]

Of course, $g$ is unique by the usual argument.

A.1.5

[definition] A morphism $f:B \to C$ is monic in $\mathcal C$ if for any distinct $e_1, e_2:A \to B$ we have $f e_1 \neq f e_2$. That is, $f$ is left-invertible.

If $B \to C$ is monic, we will sometimes say that $B$ is a subobject of $C$. [/definition]

[definition] A morphism $f:B \to C$ is epi in $\mathcal C$ if for any two distinct morphisms $g_1, g_2:C \to D$ we have $g_1 f \neq g_2 f$. That is, $f$ is right-invertible. [/definition]

[example] In $\textbf{Sets}$, $\textbf{Ab}$, and $R{-}\textbf{mod}$, the monics (resp. epi) are set injections (surjections).

This is not true in $\textbf{Rings}$ and $\textbf{Top}$. [/example]

Exercise A.1.1

[exercise] Show that $\ZZ \subseteq \QQ$ is epi in $\textbf{Rings}$. Show that $\QQ \subseteq \RR$ is epi in the category of Hausdorff topological spaces. [/exercise]

[proof=$\ZZ \subseteq \QQ$ is epi in $\textbf{Rings}$] Let $\iota:\ZZ \to \QQ$ be the inclusion map (which, of course, is a ring homomorphism). Let $f:\QQ \to R$ and $g:\QQ \to R$ be ring homomorphisms so that

Since $f\circ\iota(x) = f(x)$ and $g\circ\iota(x) = g(x)$, it follows immediately that $f = g$ on $\ZZ$. Now suppose that $a/b \in \QQ$. Then

Thus, $f = g$ on $\QQ$. [/proof]

[proof=$\QQ \subseteq \RR$ is epi in category of Hausdorff topological spaces] Now let $\iota:\QQ \to \RR$ be the inclusion map (which, of course, is continuous). Suppose that $f:\RR \to X$ and $g:\RR \to X$ are continuous maps so that

Then it follows that $f = g$ on $\QQ$. Now suppose that there is some $x \in \RR$ for which $f(x) \neq g(x)$. But $\QQ$ is dense so there exists a sequence $(x_n)\subseteq \QQ$ so that $x_n \to x$ in $\RR$. By continuity, $f(x_n) \to f(x)$ and $g(x_n) \to g(x)$. This, however, contradicts the non nonuniqueness of limits in Hausdorff spaces. [/proof]

A.1.6

[definition] An initial object (if it exists) in $\mathcal{C}$ is an object $I$ so that for every object $C \in \obj\mathcal{C}$, there is a unique morphism $I \to C$.

A terminal object (if it exists) in $\mathcal C$ is a object $T$ so that for every object $C \in \obj\mathcal C$, there exists a unique morphism $C \to T$.

An object that is both initial and terminal is called a zero object. [/definition]

Of course, all initial objects are isomorphic and all terminal objects are isomorphic.

[example] In $\textbf{Sets}$, the empty set is the initial object but it fails to be terminal in an obvious way. The terminal objects are singletons but these obviously fail to be initial.

Of course, there is a zero object in $\textbf{Ab}$ and $R{-}\textbf{mod}$. [/example]

Note that the exists of a zero object implies that each $\Hom_\mathcal{C}(B, C)$ has a distinguished element $B \to 0 \to C$ which we denote also by $0$.

[definition] Let $f:B \to C$ be a morphism. A kernel of $f$ is a morphism $i:A \to B$ so that $fi = 0$. In addition, $i$ is universal with respect to this property (i.e. any other map $e:A’ \to B$ such that $fe = 0$ must factor through $A$). [/definition]

[proposition] Every kernel is monic. [/proposition]

[proof] If $g,h:Z \to A$ so that $ig = ih$, then we have that $fig = 0 = fih$. Diagrammatrically, we are in the situation

So then $ig = ih$ must factor uniquely through $A$. Denote this morphism $\phi:Z \to A$. Then we the diagram

commutes. But since replacing $\phi$ with $g$ or $h$ still makes the diagram commute, uniqueness of $\phi$ implies $g = \phi = h$ as desired. [/proof]

[definition] Let $f:B \to C$ be a morphism. A cokernel of $f$ is a morphism $p:C \to D$ such that $pf = 0$. In addition, $p$ is universal with respect to this property. [/definition]

Since (co)kernels are universal, (co)kernels are isomorphic.

[proposition] Every cokernel is an epi. [/proposition]

Exercise A.1.2

[exercise] In $\textbf{Groups}$, show that monics are just injective set maps, and kernels are monics whose image is a normal subgroup. [/exercise]

[proof] (Injective $\implies$ monic) Obvious.

(Monic $\implies$ injective) Let $f:G \to H$ be monic with $\ker f \to G$ the obvious injection and $\phi:\ker f \to G$ the map $\phi(x) = 1$. Then $\ker f = \phi$ (as morphisms) which implies $\ker f = 1$. [/proof]

Opposite Category A.1.7

[definition] Given a category $\mathcal C$, it has an opposite category (or dual category) $\mathcal C^{\text{op}}$ where the objects are the same but composition is reversed. [/definition]

Products and Coproducts A.1.9

[definition] If $\{C_i\}_{i\in I}$ is a set of objects of $\mathcal C$, a product $\prod_{i\in I} C_i$ (if it exists)is an object of $\mathcal C$, together with maps $\pi_j:\prod C_i \to C_j$ such that for every $A \in C$, and every family of morphisms $\alpha_i:A \to C_i$, there is a unique morphism $\alpha:A \to \prod C_i$ in $\mathcal{C}$ so that $\pi_i \alpha = \alpha_i$ for all $i$.

Coproducts arise as the diagrammatic dual of a product. [/definition]

Exercise A.1.4

[exercise] Show that

and that

[/exercise]

Define a family of morphisms $\alpha_i:A\to C_i$ ($i \in I$).