1.7. Examples and Nonexamples of Functors
Functors were not defined out of arbitrary interest. The definition of a functor was motivated by constructions that were seen in mathematics (unlike constructions in say, number theory, which are interesting in their own right). Thus in this section, we include a wide variety of different constructions in in different areas of mathematics which all fit the definition of a functor. We present examples from algebraic topology, differential geometry, topology, algebraic geometry, abstract algebra and set theory.
In short, this section is due to the fact that the only way to really understand what a functor does is to realize the definition with examples. It's not necessarily important to understand all the examples, if for instance you have never worked with differential geometry, but it would be good to get a few of them. What is more important is witnessing how such a simple definition can be so versatile and prevalent in seemingly different fields of mathematics; thus, what is important is witnessing the flexibility of functors (in addition to filling in the details of the examples and doing the exercises at the end). \
{\large\noindent Algebraic Geometry.\par}
In algebraic geometry, it is often of interest to construct the affine \(n\)-space \(A^n(k)\) of a field \(k\). Usually, \(k\) is an algebraically closed field, but it doesn't have to be.
For example, when \(k = \rr\), we have that \(A^n(k) = \rr^n\). Moreover, we claim that we have a functor \(A^n(-): \fld \to \Set\). To see this, we need to figure out where \(A^n(-)\) sends objects and morphisms.
We can first observe that \(A^n(-)\) sends fields \(k\) to sets \(A^n(k)\).
Secondly, we can observe that for a field homomorphism \(\phi: k \to k'\),
we can define the function \(A^n(\phi): A^n(k) \to A^n(k')\) where
for each \((a_1, \dots, a_n) \in A^n(k)\) we have that
The reader can show that this satisfies the rest of the axioms of a functor. Overall, we can say that we have a functor
Once the affine \(n\)-space is defined, the next step in algebraic geometry is to construct the projective space \(P^n(k)\) for a field \(k\). To define this, we first define an equivalence relation on \(A^{n+1}(k)\). We say
This defines an equivalence relation on the points of \(A^n(k)\). Geometrically, this equivalence relation says two points are equivalent whenever they lie on the same line passing through the origin. With this equivalence relation, we then define
Hence we see that \(P^n(k)\) is the set of equivalence classes under this equivalence relation. Similar to the previous example, this construction is also functorial. Consider a field homomorphism \(\phi: k \to k'\). Then we define the function \(P^n(\phi): P^n(k) \to P^n(k')\) where
However, when defining functions on a set of equivalence classes, we need to be careful. It's possible that the function could send equivalent objects to different things, so that such a fuction would not be well-defined. In this case, the above function is in fact well-defined. This is because \(\phi(\lambda a_i) =\phi(\lambda) \phi(a_i)\) for any \(i = 0, 1, \dots, n\). Therefore we can state that we have a functor
\vspace{0.5cm}
{\large\noindent Algebraic Topology.\par}
An important example of a functor arises in homology theory. For example, in singular homology theory, one considers a topological space \(X\) and associates this with its \(n\)-th homology group.
In a typical topology course, one then proves that if \(f: X \to Y\) is a continuous mapping between topological spaces, then \(f\) induces a group homomorphism
in such a way that for a second mapping \(g: Y \to Z\), \(H_n(g \circ f) = H_n(g) \circ H_n(f)\) for all \(n\). Finally, we also know that \(H_n(1_X) = 1_{H_n(X)}\). Therefore, what we see is that this process can be cast into the language of category theory, so that we may define a singular homology functor
since this functorial process sends topological spaces in \(\top\) to abelian groups in \(\ab\).
Another example from algebraic topology can be realized from the fundamental group
with \(x_0 \in X\), and where \([x]\) is the equivalence class of loops based at \(x_0\), subject to the homotopy equivalence relation. First observe that \(X \mapsto \pi_1(X)\) is in fact a mapping of objects between \(\top^*\) and \(\grp\). Second, observe that if \(f: X \to Y\) is a continuous function, then we can define a group homomorphism
Note that this is well defined since if \(x \sim x'\) then there is a homotopy relation \(H: X \times [0, 1] \to Y\). However, \(f \circ H\) is also another homotopy relation that establishes that \(f(x) \sim f(x')\); hence our group homomorphism is well defined.
Moreover, if \(f: X \to Y\) and \(g:Y \to Z\) are continuous, then we can check that \(\pi_1(g \circ f) = \pi_1(g) \circ \pi_1(f)\); if \([\alpha] \in \pi_1(X, x_0)\), then
so that \((g \circ f)_* = g_*f_*\). Finally, we can examine how the identity map \(1_X\) on a topological space acts on an element \([\alpha] \in \pi_1(X, x_0)\):
so that it is sent to the identity homomorphism. All together, this allows us to conclude that this process is entirely functorial, so we may summarize our results by stating that
is a functor.
We now present two examples from differential geometry, which aren't traditionally presented as examples of functors but are nevertheless interesting in their own right. \vspace{0.5cm}
{\large\noindent Differential Geometry.\par}
Let \(M^n\) be a differentiable manifold of dimension \(n\). In general, this means that there exists a family of open sets \(U_\alpha \subset \rr^n\) and injective mappings \(\bm{x}_\alpha: U_\alpha \to M\) for \(\alpha \in \lambda\), \(\lambda\) an indexing set, with the mappings subject to various conditions\footnote{There isn't a universally agreed upon set of conditions, and we won't really need to worry about them here. If the reader likes, they can consult Do Carmo's Riemannian Geometry, which is, and has been for a long time, the go-to differential geometry text. }. Recall from differential geometry that we can associate each point \(p \in M^n\) with its tangent space \(T_p(M)\), in the following manner.
Suppose for \(\alpha' \in \lambda\) we have that \(\bm{x}_{\alpha}: U_{\alpha} \to M\) is a mapping whose image contains \(p\) (such an \(\alpha'\) must exist). Then \(T_p(M)\) has a basis
where \(\displaystyle \frac{\partial}{\partial x_i}\) is the tangent vector of the map \(\bm{c}_i: U \to M\), which simply sends \((0, \dots ,0, x_i, 0, \dots, 0)\).
Now suppose \(\phi: M_1^n \to M_2^m\) is a differentiable mapping. Recall that the differential of \(\phi\) establishes a linear transformation between the vector spaces:
Consider the category \(**DMan**^*\) whose objects are pairs \((M^n, p)\) with \(M^n\) a differentiable manifold and \(p \in M^n\). The morphism are \((\phi, p): (M_1^n,p) \to (M_2^m, q)\) with \(\phi: M_1^n \to M_2^m\) a differentiable map and \(\phi(p) = q\). Then this process may be summarized as a functor \(T_p: **DMan**_n^* \to \vect_{\rr}\) where
One can show that the identity map is sent to the identity linear transformation on \(T_p(M)\) and that the differential respects composition, so that that the association of a manifold \(M\) (with a specified point \(p \in M\)) to its tangent space \(T_p(M)\) gives rise to a functor
Consider again a differentiable manifold \(M^n\) of dimension \(n\). Recall that we may consider the tangent bundle \(TM\) of \(M\), which is the set
The set \(TM\) simply pairs each point \(p \in M^n\) with its tangent space \(T_p(M)\). However, \(TM\) is more than such a set; it inherits the structure of a differentiable manifold from \(M\) as well. In fact, it is a manifold of dimension \(2n\).
Now suppose we have a differentiable mapping \(\phi: M_1^n \to M_2^m\). Then this induces a mapping
One can show that \((\phi, d\phi): TM_1^{2n} \to TM_2^{2m}\) is a differentiable mapping between manifolds\footnote{I wanted to show this here, but it turned out to be just tedious definition-checking, so I don't think it's appropriate to include here (\textcolor{Red}{perhaps I could make/put it in an appendix...})} At this point we may guess that we have a functor \(TB: **DMan** \to **DMan**\) ("\(TB\)" for "tangent bundle") where
To check this, we first observe that \(TB(1_{M^n}) = 1_{TM^{2n}}\). Next, suppose \(\phi: M_1^n \to M_2^m\) and \(\psi: M_2^m \to M_3^p\), and observe that
Note that above in the second step, we used the fact that \(d_{\psi \circ \phi} = d_{\psi} \circ d_{\phi}\), which we know is true from the previous example. As \(TB\) respects the identity and composition, we see that we do in fact have a functor
as desired.
\vspace{0.5cm} {\large\noindent Topology.\par}
Let \(X\) be a set. Recall that we can turn \(X\) into a topological space \((X,\tau_d)\), where \(\tau^{(X)}_{d}\) is the discrete topology, so that every subset of \(X\) is an open set. We claim that this process is functorial, so that we have a functor
This is because any function \(f: X \to Y\) extends to a continuous function \(f: (X, \tau^{(X)}_{D}) \to (Y, \tau^{(Y)}_D)\) (hopefully the abuse of notation in \(f\) is forgivable here). Hence this defines a functor, although in a simpler way than we've seen in the previous examples.
Let \((X, \tau)\) be a topological space and consider any \(x_0 \in X\). Then \((X, x_0)\) forms an element of \(\top^*\). With such a space, we can consider the loop space of \((X, x_0)\) defined to be
Here \(S^1\) is the circle. As this consists of a family of continuous functions between two topological spaces, it can be endowed with the Compact Open topology to turn it into a topological space as well. Hence we claim we have a functor
To see this, one needs to first consider a morphism in \(\top^*\), which in this case is continuous function \(f: (X, x_0) \to (Y, y_0)\) such that \(f(x_0 )= y_0\). This must then correspond with a continuous function \(\Omega(f): \Omega(X) \to \Omega(Y)\). We can define this function pointwise: for each continuous \(\phi: S^1 \to X\) such that \(\phi(0) = x_0\), we have that \(\Omega(f)(\phi) = f \circ \phi: S_1 \to Y\). In this case we see that \((f \circ \phi)(0) = y_0\), and is a continuous function, so it is well-defined.
This example can be further generalized to higher loop spaces which consider continuous functions \(\phi: S^n \to X\), rather than just having \(n = 1\).
\vspace{0.5cm}
{\large\noindent Algebras, Rings, Groups.\par}
Recall that a Lie Algebra \(\mathfrak{g}\) is a vector space \(\mathfrak{g}\) (over a field \(k\)), equipped with a bilinear operation \([-, -]: \mathfrak{g}\times \mathfrak{g} \to \mathfrak{g}\) such that
- 1. \([x, y] = -[y, x]\)
- 2. \([x, [y, z]] + [y, [z, x]] + [z, [x,y]] = 0\).
Condition (2) is referred to as the Jacobi identity, and many familiar operations on vector spaces satisfy (1) and (2). For example, the cross product on vector spaces in \(\rr^3\) satisfy these properties.
Consider an associative algebra \(A\) over a field \(k\) with (associative); recall that this too has a bilinear operation \(\cdot: A \times A \to A\) with unit \(e \in A\). Then we can use \(A\) to create a Lie algebra \(L(A)\), whose (1) underlying vector space is \(A\) and (2) whose bilinear operation is \([a, b] = a \cdot b - b\cdot a\).
Now suppose \(\phi: A \to A'\) is a morphism of algebras. Then we can associate both \(A, A'\) with their Lie algebras \(L(A), L(A')\). Further, we can construct a Lie Algebra morphism \(L(\phi): L(A) \to L(A')\), using \(\phi\), by setting \(L(\phi)(a) = \phi(a)\). This is a morphism of Lie algebras since
One can then check that \(L(1_A) = 1_{L(A)}\) and \(L(\phi \circ \psi) = L(\phi)\circ L(\psi)\), so that what we have is a functor
which associates each associative algebra with its Lie algebra structure.
Let \(R\) be a commutative ring. Recall that
\(\spec(R)\) is the set of all prime ideals of \(R\). In addition, recall
that if \(\phi: R \to S\) is a ring homomorphism and
if \(P\) is a prime ideal of \(S\), then \(\phi^{-1}(P)\) is also a prime
ideal in \(R\).
This then allows us to define a functor
where on objects \(R \mapsto \spec(R)\) and on morphisms \(\phi: R \to S \mapsto \phi^*: \spec(S) \to \spec(R)\) where \(\phi^{*}(P) = \phi^{-1}(P)\).
However, we can go even deeper than this. Recall from algebraic geometry that \(\spec(R)\) can be turned into a topological space, using the Zariski topology. However, because \(\phi^{-1}(P)\) is a prime ideal whenever \(P\) is, we see that \(\phi^*: \spec(S) \to \spec(R)\) is actually a continuous function between the topological spaces. Hence we can view this as a functor
Usually this is phrased more naturally as a functor \(**Spec**: \ring \to **Sch**\) where \(**Sch**\) is the category of schemes; this is simply because schemes are isomorphic to \(\spec(R)\) for some \(R\).
Let \(G\) be a group, and \(R\) be a ring with identity. Recall from ring theory that we can form the group ring
Thus the elements are finite sums, but we have possibly infinitely many ways of adding them. Now for two elements \(\displaystyle \alpha = \sum_{g \in G}a_kg\) and \(\displaystyle \beta = \sum_{g \in G}b_gg\), we define ring addition and multiplication as
Now suppose \(\phi: G \to H\) is any group homomorphism. With that said, we claim that \(\phi\) induces a natural ring homomorphism \(\phi^{*}: R[G] \to R[H]\) between the group rings, where
Clearly this is linear and preserves scaling; less obvious is if this behaves on multiplication, although we check that below. If \(\alpha, \beta\) defined as above then
Hence we see that \(\phi^*\) is a ring homomorphism. Therefore, what we have on our hands is a functor
Possibly, your brain may wonder: it looks like we have an assignment of rings to functors.
Perhaps this process is functorial? The answer is yes, although at the moment we don't have the necessary language to describe it; we will go over this when we introduce functor categories. \vspace{0.5cm}
{\large\noindent Set Theory\par}
Consider the power set \(\mathcal{P}(X)\) on a set \(X\). Then we can create a functor \(\mathcal{P}: \Set \to \Set\) as follows.
Observe that for any set \(X\), \(\mathcal{P}(X)\) is of course another set. Therefore objects of \(\Set\) are sent to \(\Set\), as we claim.
Now let \(f: X \to Y\) be a function between two sets \(X\) and \(Y\). Then we define \(\mathcal{P}(f) : \mathcal{P}(X) \to \mathcal{P}(Y)\) to be the function where
which is clearly in \(\mathcal{P}(Y)\). Now we must show that this function respects identity and composition properties.
- Identity. Consider the identity function \(\id_X: X \to X\) on a set \(X\). Then observe that for any \(S \in \mathcal{P}X\), we have that
Therefore, \(\mathcal{P}(\id_X) = 1_{\mathcal{P}X}\) so that \(\mathcal{P}\) respects identities. * Composition. Let \(X, Y, Z\) be sets and \(f: X \to Y\) and \(g: Y \to Z\) be functions. Let \(S \in \mathcal{P}(X)\). Observe that
Therefore we see that \(\mathcal{P}(g \circ f) = \mathcal{P}(g) \circ \mathcal{P}(f),\) so that \(\mathcal{P}\) describes a functor from \(\Set\) to \(\Set\).
As we just encountered a mass of different examples of functors from different fields, one might wonder: are there other mathematical constructions which simply do not behave exactly as a functor? The answer is yes, although finding these examples is a bit tricky. The following is a well-known example, while the one after is one I haven't seen presented elsewhere. \vspace{0.5cm}
{\large\noindent Non-functor Examples.\par}
Recall from group theory that, with every group \(G\), there is an associated subgroup of \(G\) called the center:
By definition, \(Z(G)\) is an abelian group. As every group \(G\) may be associated with an abelian group \(Z(G)\), one might expect that this process is functorial. One might prematurely denote this as
However, this is not a functor, as an issue arises with the morphisms.
Consider a group homomorphism \(\phi: G \to H\). Then for this to be a functor,
we'd naturally desire a group homomorphism \(Z(\phi): Z(G) \to Z(H)\) between
the abelian groups. The only issue is that there is no consistent way to
define such a morphism from \(\phi\). The most natural way we would attempt
to achieve this is by considering the restriction, but in general
\(\phi\big|_{Z(G)}: G \to H\) does not map into \(Z(H)\).
For example, consider the Heisenberg Group
where \(R\) is a commutative ring with identity. Observe that we can create an inclusion group homomorphism \(i: H_3(R) \to \text{GL}_3(R)\). One can show that
Hence restricting the inclusion \(i: H_3(R) \to \text{GL}_3(R)\) to \(Z(H_3(R))\) results in a group homomorphism that does not even hit \(Z(\text{GL}_3(R))\) (except of course when \(a = 0\)). Thus there is not a general way to relate these two quantities in a functorial fashion.
What follows is a second example in which a process which may appear to be functorial does not turn out to be. It can, however, be adjusted to become a functor.
Let \(X\) be a set. Recall from topology that we can treat \(X\) as a topological space by associating to it the finite complement topology:
With that said, one may suppose that we have a functor \(\text{FinC}: \Set \to \top\) where \(X \mapsto (X, \tau^X_{FC})\). This would require that each function \(f:X \to Y\) extends to a continuous function \(f: (X, \tau_{FC}^X) \to (Y, \tau_{FC}^Y)\). However, for such a function to be continuous we would need that
In general, this is not true. For example suppose \(X\) is infinite and \(Y\) is finite. Then \(Y - \varnothing\) is finite, but \(X - f^{-1}(\varnothing) = X\) is infinite. Hence this cannot define a functor \(F: \Set \to \top\).
{\large Exercises \vspace{0.5cm}}
- *1.*
- (i.) Let \(X\) and \(Y\) be two sets. Regard each set as a discrete category. Interpret what a functor \(F: X \to Y\) means in this case.
- (ii.) Let \(G\) and \(H\) be two groups. Regard each group as a one-object category whose morphisms sets correspond to their group elements, with composition their group product. Interpret what a functor \(F: G \to H\) means in this case.
- (iii.) Let \(X\) and \(Y\) be a pair of thin categories. Interpret what a functor \(F: X \to Y\) means in this case. (Use (i) to get you started.)
- *2. Let \(G\) be a group. Then for any two elements \(a, b \in G\), we define the commutator* of \(a, b\) to be the element
Define \([G, G]\) to be the set
which we call the commutator subgroup. Its underlying set consists of all possible products, with factors that are of the form \(a_ib_ia_i^{-1}b_i^{-1}\). One can show that \([G, G] \normal G\) for any group \(G\), which implies that we may discuss the quotient group \(G/[G, G]\), which is abelian in this case.
So, show that we have a functor \(F: \grp \to \ab\) where
Deduce the action of \(F\) on the morphism of \(\grp\) (i.e., the group homomorphisms.) and show that it is well-defined. \vspace{0.2cm} * *3.* Let \(R\) be a unital ring. Recall that \(GL_n(R)\) is the group consisting of \(n \times n\) matrices with entries in \(K\). Show that this construction more generally is that of a functor
In addition, with such a ring \(R\), we may associate it with its group of units \(R^{\times}\), which you may recall is
Show that this also defines a functor
We will see in the next section that there is an interesting relationship between these two functors. * *4.* Recall the category \(\Set_{FTO}\) is the category whose objects are sets and whose morphisms are functions with the finite-to-one property (See Exercise 1.3.3). While we saw that \(\text{FinC}: \Set \to \top\) where
does not define a functor, show that upon changing the domain
category from \(\Set\) to \(\Set_{FTO}\), it does
define a functor \(\text{FinC}: \Set_{FTO} \to \top\).
\vspace{0.2cm}
* *5.
* (i.)* Let \(X = \{x_1, x_2, \dots, x_n\}\) be a finite set. With such
a finite set, we can
pick a field \(k\) and build \(X\) into a finite-dimensional
vector space \(V_X\) over \(k\). Explicitly, we can create the vector
space
$$
V_X= \{c_1x_1 + \cdots + c_nx_n \mid c_i \in k\}.
$$
We define addition in the intuitive way of adding coefficients of the
same basis, so this is truly a vector space. Note that when $k = \rr$,
we get that $V_X \cong \rr^n$.
Prove that this process is functorial. That is, show that the functor
$$
F: \finset \to \vect_k \qquad F(X) = V_X
$$
is a functor.
* **(*ii*).** From any set $X$, we may construct the **free group** $F(X)$ generated
by $X$. The elements of $F(X)$ are (1) the elements of $X$, (2) a new
element $e$, and (3) all elements $xy$ whenever $x, y \in X$.
In this way, $F(X)$ is a group with the product being string concatenation,
and we require that $xe = x = ex$.
. Below, two words (elements of $F(X)$) are combined.
$$
(x^2yz^{-1}) \cdot (zy^2x) = x^2y^2x.
$$
Show that we have a functor $F: \Set \to \grp$ where
sets are mapped to their free groups, that is, $X \mapsto F(X)$.
* **(*iii*).** For any set $X$, we can build the **free ring** $(R\{X\}, +, \cdot)$
as follows. Let $(F(X), \cdot)$ be the free group with the added relation that
$xy = yx$ for any $x, y \in F(X)$. We can then define
$$
R\{X\} = \left\{ \sum_{x_i \in F(X)} x_i^{n_i} \mid \right\}
$$
- *6. Let \(V\) be a vector space over a field \(k\). Recall that we can associate \(V\) with its projective space* \(P(V)\) which is defined as the set of equivalence classes of element in \(V\), subject to the relation \(v \sim w\) if \(v = \lambda w\) for some nonzero \(\lambda \in k\). That is,
where \([v]\) denotes the equivalence class of \(v\). Show that this process is functorial, so that we have a functor
$$ P: \vect_k \to \Set. $$ * *7. Let \(R\) be a ring with ideal \(I\). Recall that we can construct the radical of the ideal \(I\)* as the ideal
Show that we have a functor
where \(**Ideals**(R)\) is the partial order of ideals on \(R\), whose ordering is given by subset containment. * *8.* Let \(X\) be a topological space, and denote \(\open(X)\) as the category where the objects are open sets \(U \subset X\) and morphisms are inclusion morphisms. Create a functor
where on objects \(F(U) = \{f: U \to \rr \mid f \text{ is continuous}\}\). That is, how should \(F\) act on the morphisms for this to be a functor? * *9.* Let \(k\) be a field. With each field, we may associate \(k\) with the category \(\vect_k\) which consists of finite dimensional vector spaces \(V\) over \(k\). Is this process functorial? That is, do we have a functor
where \(\vect(k) = \vect_k\)?\ Hint: No. But explain why it breaks.