1.1Basic Topology

A category $\C$ comprises:

1. Class of Objects. a class of objects $\Obj(\C)$.

2. Set of Morphisms. for any two objects $X, Y \in \Obj(\C)$, a set of morphisms $\C(X, Y)$ whose elements are maps $f: X \to Y$.

3. Law of Composition. a composition rule such that, for any two morphisms $f: X \to Y$ and $g: Y \to Z$, there exists a morphism $gf: X \to Z$.

A category obeys the following rules:

1. Associativity of Composition. That is, for morphisms $f: X \to Y$, $g: Y \to Z$, and $h: Z \to A$, we have that $(hg)f = h(gf)$.

2. Identity Morphisms. For any two objects $X, Y \in \Obj(\C)$ and a morphism $f: X \to Y$, there exist identity morphisms $\id_X: X \to X$ and $\id_Y: Y \to Y$ such that $f \id_X = f = \id_Y f$.

Let $\C$ be a category, and let $X, Y \in \Obj(\C)$ be arbitrary objects. Then $f: X \to Y$ is said to be left-invertible (resp. right-invertible) if there exists a morphism $g: Y \to X$ such that $gf = \id_X$ (resp. $h: Y \to X$ such that $fh = \id_Y$).

In the case where $f: X \to Y$ has both a left inverse and right inverse, then

and the single morphism $g = h$ is said to be the inverse of $f$. A morphism $f: X \to Y$ is said to be an isomorphism if it is invertible, and the objects are said to be isomorphic (denoted $X \cong Y$).

An isomorphism is a type of equivalence relation—given that it is reflexive, symmetric, and transitive—and so it forms equivalence classes, called isomorphism classes. For example, isomorphisms in $\Set$ are bijections, and two isomorphic sets are said to have the same cardinality; a cardinal is an isomorphism class of sets.

Let $\C$ be a category with three objects $X, Y, Z \in \Obj(\C)$, and let $f \in \C(X, Y)$. The pushforward of $f$ is the map of sets $f_*: \C(Z, X) \to \C(Z, Y)$ such that $f_*: g \mapsto fg$. The pullback of $f$ is the map of sets $f^*: \C(Y, Z) \to \C(X, Z)$ such that $f^*: g \mapsto gf$.

Consider the same setup as in the previous definition. For some $Z \in \Obj(\C)$ and some morphism $f: X \to Y$, $f$ induces a pushforward on maps from $Z$ to $X$. Namely, if $g: Z \to X$ is a morphism, we can apply $f$ so that $fg: Z \to Y$, and so the result is pushed to $Y$ from $X$. On the other hand, $f$ also induces a pullback on maps from $Y$ to $Z$. Namely, if $g: Y \to Z$, we can apply $f$ so that $gf: X \to Z$, and so the inputs are pulled back into $X$ from $Y$.

The following are equivalent.

1. $f: X \to Y$ is an isomorphism.

2. The pushforward $f_*: \C(Z, X) \to \C(Z, Y)$ is an isomorphism of sets.

3. The pullback $f^*: \C(Y, Z) \to \C(X, Z)$ is an isomorphism of sets.

A functor $F$ is a mapping from a category $\C$ to a category $\sf{D}$ that ensures the following data.

1. For every $X \in \Obj(\C)$, there is some $FX \in \Obj(\mathsf{D})$.

2. For any two $X, Y \in \Obj(\C)$ and any $f \in \C(X, Y)$, there is some $Ff \in \mathsf{D}(FX, FY)$.

A functor must obey the following properties.

1. $(Fg)(Ff) = F(gf)$ for any morphisms $f: X \to Y$ and $g: Y \to Z$.

2. $F(\id_X) = \id_{FX}$ for any $X \in \Obj(\C)$.

A functor $F$ is said to be covariant if it maps the category $\C$ into $\mathsf{D}$ as defined above. A functor $F$ is said to be contravariant if it maps the category $\C^\op$ into $\mathsf{D}$. Note that $\C$ and $\C^\op$ have the same objects, so the first rule doesn't change, but the arrows are reversed, so every morphism $f: X \to Y$ in $C$ is assigned a morphism $Ff: FY \to FX$ in $D$.

Consider the category of vector spaces over a field $k$, whose objects are vector spaces over $k$ and whose morphisms are linear maps $T: V \to W$. Consider also the category of dual vector spaces, where objects are of the form $\Hom(V, k)$ for some vector space $V$ and morphisms are dual maps $T^*: W^* \to V^*$. Then consider a functor $F$ between these two categories, one that sends $V \mapsto \Hom(V, k)$ and $T \mapsto T^*$. This is a contravariant functor: a morphism in the category of vector spaces is a linear map from $V$ to $W$, but after the functor is applied, it becomes a map from $W^* = FW$ to $V^* = FV$, an obvious reversal of the morphism's direction.

1. Consider the functor $F = \C(X, -)$. When given a functor in this notation, objects $Z \in \Obj(\C)$ are mapped to an object by replacing $-$ with the object in question, and so $FZ = \C(X, Z)$. Morphisms $f: Y \to Z$ are mapped to a morphism by replacing the $-$ with the domain and codomain, and so $Ff : \C(X, Y) \to \C(X, Z)$, which is exactly the pushforward $Ff = f_*$.

2. The contravariant functor $\C(-, X)$ from $\C^\op$ to $\Set$ maps objects in $\C$ to the set of all morphisms from itself to $X$, and it maps morphisms to their pullbacks. Indeed, if $Y \in \Obj(\C)$, then the functor acts by $Y \mapsto \C(Y, X)$, taking $Y$ to the set of all morphisms from $Y$ to $X$. If $f: Y \to Z$ is a morphism, note that by contravariance it is mapped to a morphism $\C(Z, X) \to \C(Y, X)$, which is exactly the pullback.

3. The functor $\Hom(-, k)$, where inputs are taken from the category of vector spaces, maps each vector space to its dual and maps each linear map to its dual map.

4. A forgetful functor acts by forgetting the special features and structure of the domain. For example, a forgetful functor from $\Top$ to $\Set$ works by mapping all objects (topological spaces) in $\Top$ to a set without its topology, and it can map all morphisms (continuous maps) in $\Top$ to just a general function.

5. A free functor is conceptually the “opposite” of a forgetful functor; it adds structure to a set in the most generic, canonical way possible. For example, a free functor from $\Set$ to $\mathsf{Grp}$ takes a set of generators $S$ and maps it to the free group $F(S)$. We omit the formal definition for now, as it requires the machinery of adjoint functors (to be discussed later).

Let $F$ be a functor from category $\C$ to category $\mathsf{D}$. Then the functor induces a map between hom-sets in $\C$ and $\mathsf{D}$, defined by

for any $X, Y \in \Obj(\C)$. If the map is

1. injective, $F$ is called a faithful functor.

2. surjective, $F$ is called a full functor.

3. bijective, $F$ is called a fully faithful functor.

Suppose $F: \C \to \mathsf{D}$ is a functor. If $X \cong Y$ in $\mathsf{C}$, then $FX \cong FY$ in $\mathsf{D}$.

If two objects $X, Y \in \Obj(\C)$ are “the same”, then two objects $FX, FY \in \Obj(\mathsf{D})$ are also “the same.” We say $FX$ is an invariant of $X$. We say $F$ computes the invariant for $X$: for any data shared between isomorphic objects in $\C$, the functor encodes this in $\mathsf{D}$. For example, consider the forgetful functor from $\Top$ to $\Set$: this functor acts as an invariant since if any two topological spaces are homeomorphic, their sets are bijective. We may also consider the contrapositive, that if $FX \not \cong FY$, then $X \not \cong Y$. Once again, using the forgetful functor, one notes that if the underlying sets of topological spaces are not bijective, then they are certainly not homeomorphic.

Suppose $F, G$ are two functors from $\C$ to $\mathsf{D}$. A natural transformation $\eta$ from $F$ to $G$ comprises morphisms $\eta_X: FX \to GX$ for each $X \in \Obj(\C)$, where these morphisms make the diagram

commute (i.e., $\eta_Y Ff = Gf \eta_X$). If $\eta_X: FX \to GX$ is an isomorphism for each $X \in \Obj(\C)$, we say $\eta$ is a natural isomorphism/natural equivalence, and we say $F$ and $G$ are naturally isomorphic, denoted $F \cong G$. The collection of all natural transformations between two functors is denoted $\Nat(F, G)$.

For every object $X$ in the category $\C$ and for every contravariant functor $F: C^\op \to \Set$, we have that

That is, every element of $FX$ is in bijection with a natural transformation $\eta: C(-, X) \to F$.

We omit the proof (it follows readily from the diagram for natural transformations), but consider the special case of the contravariant functor $C(-, Y)$. Then we have that

For any categories $C, D$, let $D^C$ denote the category whose objects are functors $F: C \to D$ and whose morphisms are natural transformations $\eta: F \to G$. Now let $D = \Set$, and consider the category $D^{C^{\op}} = \Set^{C^\op}$. Note that the objects of this category are contravariant functors $F: C^\op \to \Set$, and a functor like this is called a presheaf. This category itself is quite important since it has nice properties, but for now we will discuss a special functor.

Note that $C(-, X)$ is a contravariant functor from $C^\op$ to $\Set$, and so it is an object in our category $\Set^{C^\op}$. Let us consider the functor $y: C \to \Set^{C^\op}$, which sends objects $X \mapsto C(-, X)$ and sends morphisms $f: X \to Y$ to a natural transformation $\eta$ whose components are all pushforwards of $f$. That is, for each $Z \in \Obj(C)$, we have that $\eta_Z: C(Z, X) \to C(Z, Y)$, which is exactly the pushforward of $f$ when considering a map $g: Z \to X$.

Consider the map $f \mapsto yf$, which is a map $C(X, Y) \to \Set^{C^\op}(FX, FY)$, which is equivalently rewritten as $C(X, Y) \to \Nat(C(-, X), C(-, Y))$. Note that $\Nat(C(-, X), C(-, Y)) \cong C(X, Y)$ by Yoneda Lemma. We can show that $f$ is the inverse of the bijection $\Phi(\eta) = \eta_X(\id_X)$, and so $f$ is a bijection.

Suppose $f: X \to Y$ is a morphism. We say $f$ is a monomorphism if $f$ is left cancellative. That is, for any two morphisms $g_1, g_2: Z \to X$, we have that

Conversely, we say $f$ is an epimorphism if $f$ is right cancellative. That is, for any two morphisms $g_1, g_2: Y \to Z$, we have that

We denote a monomorphism by $\mono$ and an epimorphism by $\epi$.

Note that monomorphisms coincide with injections and epimorphisms coincide with surjections in $\Set$.

If $f$ is an injective function and $f(g_1(x)) = f(g_2(x))$, it is clear that $g_1(x) = g_2(x)$ by injectivity, and so $f$ is left cancellative (and thus monic). On the other hand, we can prove that if $f$ is left cancellative, $f$ is injective by contraposition. If $f$ is not injective, then there exists $x_1 \ne x_2$ for which $f(x_1) = f(x_2)$. Let $g_1, g_2$ be maps such that $g_k(x) = x_k$, and so $f(g_1(x)) = f(g_1(x))$ but $g_1(x) \ne g_2(x)$, and so $f$ is not left cancellative. Thus a map in $\Set$ is injective if and only if it is monic.

If $f$ is a surjective function and $g_1(f(x)) = g_2(f(x))$ for every $x$, then since every $y$ in the codomain is some $f(x)$, we have $g_1(y) = g_2(y)$ for every $y$. Thus $f$ is right cancellative (and thus epic). Conversely, we show that if $f$ is not surjective, it is not epic. If $f$ is not surjective, there is some $y_0 \in Y$ not in the image of $f$. Define $g_1, g_2: Y \to \{0, 1\}$ such that both maps agree on the image of $f$ (e.g., they both map every $y \in f(X)$ to $1$), but $g_1(y_0) = 1$ and $g_2(y_0) = 0$. Then $g_1 f = g_2 f$ but $g_1 \neq g_2$, so $f$ is not right cancellative.

A map that is both monic and epic in $\Set$ is also an isomorphism, since it forces a unique inverse for the map. This is not always true in other categories. For example, a continuous bijection in $\Top$ is monic and epic, but it is not a homeomorphism (and thus not an isomorphism).

Suppose $\C$ is a category. An object $X \in \Obj(\C)$ is called initial if there is a unique map from $X$ to every other object in $\C$. Conversely, $X$ is called final if there is a unique map from every other object in $\C$ to $X$.

The empty set is the initial object in $\Set$. There is exactly one map $f: \{\} \to X$ for any set $X$, which is the “empty map”, or in other words the map that does nothing. Conversely, all singletons $\{*\}$ are terminal in $\Set$. For any set $X$, there is exactly one map $f: X \to \{*\}$, which is the constant map $f(x) = *$ for every $x \in X$. Note that singletons are unique up to unique isomorphism in $\Set$, abiding by our previous intuition.

Recall the Cartesian product of two sets $X \times Y$ is the set of all ordered pairs $(x, y)$, with $x \in X$ and $y \in Y$. We have a concrete characterization of what the structure inside a Cartesian product looks like, but we want a more complete characterization. We want to be able to describe morphisms into or out of $X \times Y$. We can do this by interpreting the product categorically.

We form a product $X \times Y$ by taking two sets $X, Y$ together with two projection maps $\pi_X(x, y) = x$ and $\pi_Y(x, y) = y$. We characterize the product of two sets by the property that if we have any two maps $f_X: Z \to X$ and $f_Y: Z \to Y$, we can factor these maps through the product by $f_X = \pi_X \circ h$ and $f_Y = \pi_Y \circ h$, where $h: Z \to X \times Y$ is unique. We can represent this with a commutative diagram.

Note that $h$ forces this diagram to commute. If we drew $f_X, f_Y$ out of $Z$ and $\pi_X, \pi_Y$ out of $X \times Y$, the diagram wouldn't be “connected” (thus wouldn't commute) until a map was drawn from $Z \to X \times Y$. The universal property of a product completely characterizes every single map going into $X \times Y$ from some set by the relationships $\pi_X \circ h = f_X$ and $\pi_Y \circ h = f_Y$. Using this, we have a straight forward proof that all the sets that satisfy this universal property are unique up to unique isomorphism, and in the language of category theory, we have that $X \times Y$ is well defined.

Note that we can extend the definition of a product from a product of two objects to an arbitrary product by an indexing set $\Lambda$, simply by assigning the product a projection map $\pi_\lambda$ for each $\lambda \in \Lambda$. This is one nice aspect of the categorical product: it scales nicely. Its main utility, however, is characterizing products in different categories. If a construction in some other category satisfies this universal property, we have established a concrete link between the product in $\Set$ with whatever construction is described by the universal property in the other category.

Recall the coproduct of two sets $X \sqcup Y$ (often called the disjoint union) is the set of all elements belonging to either $X$ or $Y$, typically distinguished by indexing tags: $X \sqcup Y = (X \times \{0\}) \cup (Y \times \{1\})$. We have a concrete characterization of what the structure inside a disjoint union looks like, but we want a more complete characterization. We want to be able to describe morphisms into or out of $X \sqcup Y$. We can do this by interpreting the coproduct categorically.

We form a coproduct $X \sqcup Y$ by taking two sets $X, Y$ together with two canonical injection maps $i_X(x) = (x, 0)$ and $i_Y(y) = (y, 1)$. We characterize the coproduct of two sets by the property that if we have any two maps $f_X: X \to Z$ and $f_Y: Y \to Z$, we can factor these maps through the coproduct by $f_X = h \circ i_X$ and $f_Y = h \circ i_Y$, where $h: X \sqcup Y \to Z$ is unique. We can represent this with a commutative diagram.

Note that $h$ forces this diagram to commute. If we drew $f_X, f_Y$ into $Z$ and $i_X, i_Y$ into $X \sqcup Y$, the diagram wouldn't be “connected” (thus wouldn't commute) until a map was drawn from $X \sqcup Y \to Z$. The universal property of a coproduct completely characterizes every single map going out of $X \sqcup Y$ to some set by the relationships $h \circ i_X = f_X$ and $h \circ i_Y = f_Y$. Using this, we have a straightforward proof that all the sets that satisfy this universal property are unique up to unique isomorphism, and in the language of category theory, we have that $X \sqcup Y$ is well defined.

Note that we can extend the definition of a coproduct from a coproduct of two objects to an arbitrary coproduct by an indexing set $\Lambda$, simply by assigning the coproduct an injection map $i_\lambda$ for each $\lambda \in \Lambda$. This is one nice aspect of the categorical coproduct: it scales nicely. Its main utility, however, is characterizing coproducts in different categories. If a construction in some other category satisfies this universal property, we have established a concrete link between the coproduct in $\Set$ with whatever construction is described by the universal property in the other category.