Degree and Euclidean division of polynomials

For the definition and characterization of the polynomials over a field \K, see Polynomials over a field.

Definition of the degree of a polynomial

The degree of a polynomial P, written \deg P, is an element of \{-\infty\} \cup \N. The symbol -\infty is taken to have the usual properties of order and of addition relative to natural numbers.

The family (X^0, X^1, X^2, \ldots) is a \K-vector space basis of \K[X]. For any P \in \K[X], there exists an unique (a_0, a_1, \ldots) \in \K^\N, only finitely many of which are nonzero, such that P = \sum_n a_n X^n. If all the a_n are zero, that is, if P is the zero element of \K[X], then \deg P is defined as -\infty. If not all the a_n are zero, since only finitely many are nonzero, there is a greatest value of n such that a_n \neq 0_\K. Then \deg P is defined as this greatest value.

Elementary properties of the degree

The following are easily checked to hold for any polynomials P and Q (even when one or both are zero):

\deg (P + Q) \le \sup(\deg P, \deg Q)

if \deg P \neq \deg Q, then \deg (P + Q) = \sup(\deg P, \deg Q)

\deg PQ = \deg P + \deg Q

Euclidean division

The notion of the degree of a polynomial allows us to formulate the following property of Euclidean division:

For all polynomials A and B, with B nonzero, there exists exactly one pair of polynomials (Q, R) such that A = Q B + R with \deg R < \deg B.

For the proof, we fix the value of nonzero polynomial B, and recurse over the degree of A.

More specifically, B being given, we consider the following proposition dependent on n \in \{-\infty\} \cup \N:

\mathcal P(n) iff for all polynomials A such that \deg A \le n, there exist polynomials  Q, R, with \deg R < \deg B, such that A = Q B + R.

\mathcal P(-\infty) is trivially true (for \deg A \le -\infty means that A is zero, and Q = 0, R = 0 satisfies A = Q B + R with \deg R < \deg B).

Let us suppose \mathcal P(n) true for some n \in \{-\infty\} \cup \N, and consider a polynomial A the degree of which is less or equal to the successor of n, that is to 0 if n = -\infty, or n+1 otherwise.

If the degree of A is less or equal to n, then \mathcal P(n) applies and there exist the wanted Q and R.

If the degree of A is less than that of B, then we can directly write A = Q B + R with Q = 0 and R = A.

Remains the case in which \deg A is the successor of n.

Polynomials over a field

In this context, we consider a commutative field (simply: field) \K. The polynomials will be constructed as a certain associative unital algebra over \K (“\K-aua”), together with a distinguished element called the formal variable.

\K itself, as a vector space over itself and the usual multiplication in \K as the third law, is a \K-aua,

Morphisms between two \K-auas are linear mappings that conserve the third law and map the unit of the first algebra onto the unit of the second one.

All that will be said of the polynomials over \K will be based on the universal property given below. The polynomials will thus be defined uniquely up to an isomorphism.

Definition

The polynomials over \K is any \K-aua A together with a distinguished element X of A such that the following universal property holds:

For any \K-aua A' and any a \in A, there exists exactly one morphism \phi: A \to A' such that \phi(X) = a.

Unicity up to an isomorphism

Let (A_1, X_1) and (A_2, X_2) be two sets of polynomials over \K. The universal property implies the existence of a morphism \phi_1: A_1 \to A_2 such that \phi(X_1) = X_2, and of a morphism \phi_2: A_2 \to A_1 such that \phi(X_2) = X_1.

We then have (\phi_2 \circ \phi_1)(X_1) = X_1. Hence \phi_2 \circ \phi_1 is a morphism A_1 to A_1 that maps X_1 to X_1. Now \Id_{A_1} is another such morphism. By virtue of the universal property of (A_1, X_1), it follows that \phi_2 \circ \phi_1 = \Id_{A_1}, that is, is the \K-aua category identity on A_1.

The same reasoning shows that \phi_1 \circ \phi_2 is the \K-aua category identity on A_2.

Hence \phi_1 is an isomorphism A_1 \to A_2 that maps the formal variable of A_1 to that of A_2.

Characterization

Let A be an associative unital algebra over \K, and X an element of A. Then (A, X) represents the polynomials over \K if and only if the family (X^n)_{n \in \N}, with X^0 = 1_A and X^{n+1} =  X X^n, is a family-base of the \K-vector space A.

Let us first suppose that the latter condition holds on (A, X), and prove the universal property.

Let A' be a \K-aua and a' an element of A'.

Let us suppose that \phi is a morphism A \to A' such that \phi(X) = a'. Then for all n \in \N, we have \phi(X^n) = a'^n. Since \phi as a \K-aua morphism, it is in particular a linear mapping. Since a linear mapping is defined by the image of a basis, and (X^n)_{n \in \N} is a basis, \phi necessarily is the unique linear mapping A \to A' such that \forall n \in \N, \phi(X^n) = a'^n. Now it is easy to check that this linear mapping does preserve the multiplication in A, and is hence a \K-aua morphism.

Thus the universal property is proven for (A, X).

Let us now suppose the universal property for (A, X), and show that (X^n)_{n \in \N} is a family-base of the \K-vector space A.

Let us first show that (X^n)_{n \in \N} is linearly independent.

Let (A', \alpha) be a free \K-vector space on set \N. This entails that (\alpha(n))_{n \in \N} is a family-base of A'. To make A' into an associative unital algebra, we must define a multiplication law that is bilinear, and also associative.

A bilinear law can be defined by the images of all the ordered pairs of base elements. Hence we may define the multiplication in A' by:

\forall n_1, n_2 \in \N, \alpha(n_1) \cdot \alpha(n_2) = \alpha(n_1 + n_2)

We then have (\alpha(n_1) \cdot \alpha(n_2)) \cdot \alpha(n_3) = \alpha(n_1 + n_2) \cdot \alpha(n_3) = \alpha(n_1 + n_2 + n_3) = \alpha(n_1) \cdot \alpha(n_2 + n_3) = \alpha(n_1) \cdot (\alpha(n_2) \cdot \alpha(n_3)).

Hence the multiplication is associative with regard to the elements of the base; it is easy to check that this follow through to arbitrary elements, which makes A' an associative algebra.

\alpha(0) is a unit; hence A' is an associative unital algebra.

Applying the universal property of (A, X), we find that there exists exactly one \K-aua morphism \phi: A \to A' such that \phi(X) = \alpha(1).

We then have \phi(X^0) = \phi(1_A) = 1_{A'} = \alpha(0); \phi(X^1) = \phi(X) = \alpha(1); and, for any n > 1, \phi(X^n) = \phi(X \cdot \ldots \cdot X) = \phi(X) \cdot \ldots \cdot \phi(X) = \alpha(1) \cdot \ldots \cdot \alpha(1) = \alpha(1 + \ldots + 1) = \alpha(n). Hence for all n \in \N, \phi(X^n) = \alpha(n).

Let us consider a linear combination of the family (X^n)_{n \in \N} that vanishes; that is, a finite sequence of values of \K, (a_0, a_1, \ldots a_n), such that \sum_{i = 0}^n a_i X^i = 0_A. Taking the image of this linear combination by \phi, we obtain \sum_{i = 0}^n a_i \alpha(i) = 0_{A'}. Since the \alpha(i) form a basis, all the a_i must be zero.

Thus the family (X^n)_{n \in \N} is linearly independent.

Lastly, we must show that this family generates the \K-vector space A.

Let B be the sub-aua of A generated by \{X\}.

There exists a unique aua morphism \phi: A \to B such that \phi(X) = X.

Let then \phi' be \phi, but with all of A as codomain; in other words, \phi': A \to A, x \mapsto \phi(x). It is clear that \phi' too is an aua morphism, and it is such that \phi'(X) = X.

But there is exactly one aua morphism A \to A that maps X to X; and \Id_A is such.

Hence \phi' = \Id_A.

The image of \phi' is that of \phi, and is a subset of B. But the image of \Id_A is A itself. Hence B = A.

Thus the sub-aua of A generated by \{X\} is A itself. This sub-aua is the set of linear combinations of all the powers of X; hence the sub-vector space generated by X^0, X^1, X^2, \ldots is A.

Thus (X^n)_{n \in \N} is a family-base of the \K-vector space A.

Equivalent alternative definitions concerning filters

Definition of a filter

Let E be a set.

Traditional definition of a filter on E: A collection \mathcal F of subsets of E is a filter on E iff \emptyset \notin E; and \forall A \in \mathcal F, \forall B \subseteq E, (B \supseteq A \implies B \in \mathcal F); and \forall A, B \in \mathcal F, A \cap B \in \mathcal F.

Alternative definition: A collection \mathcal F of subsets of E is a filter on E iff \emptyset \notin E; and \forall A, B \subseteq E, ((A, B) \in \mathcal F^2 \iff A \cap B \in \mathcal F).

The two definitions are equivalent.

Definition of the image of a filter

Let E and F be sets, \mathcal F a filter on E and f: E \to F.

Traditional definition: The filter on F image of \mathcal F by f is (f^{\rightarrow \rightarrow}(\mathcal F))^{\uparrow F}.

Alternative definition: The filter on F image of \mathcal F by f is \{\ B \subseteq F\ |\ f^\leftarrow(B) \in \mathcal F\ \}.

Again, the two definitions are equivalent.

Margin between a compact subspace and an including open subspace

In Chapter 31 (“Compact-Open Topology”), the fact that, on the set of the continuous mappings \R \to \R the second topology (the uniform convergence topology) is finer than the first (the compact-open topology) is said to be “clear” but actually needs some non-trivial proof. Part of this is the lemma that I state and prove below.

For any x \in E and real r > 0, let \dot B(x, r) be the open ball centered on x and with radius r. We know that an open ball is an open subset.

The lemma says:

Let (E, d) be a metric space, O an open subspace of E and C \subseteq O a compact subspace of E. Then there exists r > 0 such that for all x in C, \dot B(x, r) \subseteq O.

Proof:

For each x \in C, since x \in O, there exists an r_x > 0 such that \dot B(x, r_x) \subseteq O.

The collection of all \dot B(x, \frac 1 2 {r_x}) with x \in C is an open cover of C. Hence there exists a finite J \subseteq C such that \{\ \dot B(x, \frac 1 2 {r_x})\ \}_{x \in J} covers C.

Since J is finite, there exists a real r > 0 such that for all x \in J, r \le \frac 1 2 {r_x}.

Let x be an eventual point of C.

Since \{\ \dot B(z, \frac 1 2 {r_z})\ \}_{z \in J} covers C, there exists z \in C such that x \in \dot B(z, \frac 1 2 {r_z}), that is, d(z, x) < \frac 1 2 {r_z}.

Let y be an eventual point of \dot B(x, r).

We have d(z, x) < \frac 1 2 {r_z} and d(x, y) < r \le \frac 1 2 {r_z}, which leads to d(z, y) < r_z, that is, y \in \dot B(z, r_z). But z is in C and r_z was taken such that \dot B(z, r_z) \subseteq O. Hence y \in O.

Hence \dot B(x, r) \subseteq O.

We thus have r > 0 such that \forall x \in C,\ \dot B(x, r) \subseteq O. \blacksquare

Open tubes

In Chapter 31, “The Compact-Open Topology”, it is asserted that “clearly” the third topology on \Mor(X, Y) (with X and Y the real line) is finer than the second. The issue is more fully discussed here.

In this post I discuss the concept of open tubes, particular subsets of the Cartesian product X \times Y with X a set and Y a metric space.

Definition

Let X be a set and Y a metric space, with distance d.

For any r > 0 and any a \in Y, we note \dot B(a, r) the open ball of Y of radius r and centred on a, that is set of all x \in Y such that d(a, x) < r. Open balls are notoriously open subsets.

Let \phi be any mapping X \to Y, and r a real > 0.

Then the open tube \dot T(\phi, r) of radius r and centred on \phi is the following subset of X \times Y:

\dot T(\phi, r) = \{\ (x, y) \in X \times Y\ |\ d(\phi(x), y) < r\ \}

Equivalently:

\dot T(\phi, r) = \bigcup_{x \in X} \{x\} \times \dot B(\phi(x), r)

The “open” in “open tube” refers to its being based on open balls (second expression above), but does not imply that it is itself necessarily open; indeed, we have not given any topology to X, so we don’t have any particular one on X \times Y and cannot speak of \dot T(\phi, r) being open or not.

However, if X is a topological space and \phi is continuous X \to Y, then \dot T(\phi, r) is open, as I will show.

An open tube centred on a continuous mapping is open

If X is a topological space and \phi is a continuous mapping X \to Y, then any open tube centred on \phi is an open subset of X \times Y.

Proof:

I will show that in the above circumstances \dot T(\phi, r) is a neighbourhood of each of its points.

Let (x_0, y_0) be an eventual point of \dot T(\phi, r).

Then d(\phi(x_0), y_0) < r.

Lettuce define s = \frac 1 2 (r - d(\phi(x_0), y_0)). We have s > 0.

Let D = \phi^\leftarrow(\dot B(\phi(x_0), s)). Since \phi is continuous and \dot B(\phi(x_0), s) is an open subset of Y, D is an open subset of X. Since \phi(x_0) \in \dot B(\phi(x_0), s), x_0 \in D.

Let O = D \times \dot B(y_0, s), open subset of X \times Y. Clearly, (x_0, y_0) \in O, so O is a neighbourhood of (x_0, y_0).

Let (x, y) be an eventual point of O.

x \in D, hence \phi(x) \in \dot B(\phi(x_0), s), that is d(\phi(x), \phi(x_0)) < s.

y \in \dot B(y_0, s), hence d(y_0, y) < s.

d(\phi(x), y) \le d(\phi(x), \phi(x_0)) + d(\phi(x_0), y_0) + d(y_0, y)

Since d(\phi(x), \phi(x_0)) < s and d(y_0, y) < s, this leads to:

d(\phi(x), y) < s + d(\phi(x_0), y_0) + s

With:

s + d(\phi(x_0), y_0) + s = 2 s + d(\phi(x_0), y_0)
\ = 2 (\frac 1 2 (r - d(\phi(x_0), y_0))) + d(\phi(x_0), y_0)
\ = r

Thus we have: d(\phi(x), y) < r.

Which implies: (x, y) \in \dot T(\phi, r).

Hence O \subseteq \dot T(\phi, r).

Thus we have found a neighbourhood O of (x_0, y_0) that is a subset of \dot T(\phi, r), which implies that \dot T(\phi, r) itself is a neighbourhood of (x_0, y_0).

Thus \dot T(\phi, r) is an open subset of X \times Y. \blacksquare

Uniform convergence topology and the open set topology

Chapter 31, “The Compact-Open Topology”, describes the compact-open topology, and goes on to compare it with two other topologies on \Mor(X, Y) in the case where X = Y = \R. It asserts that “clearly” the second of these three is finer than the first, and the third finer than the second.

This may well be “clear” intuitively, but is not so easy to prove. Here I will prove that the third (that the author does not name, but that I call the open set topology) is finer than the second, which is the uniform convergence topology.

I will only need to specify that X is a topological space and Y a metric space with distance d.

I define E = \Mor(X, Y) the set of continuous mappings from X to Y.

I note \dot B(y, r) the open ball in Y centred on y \in Y and with radius r > 0. It is the set of all y' \in Y such that d(y, y') < r.

Furthermore, for any H subset of X \times Y, I note \kappa(H) the set of all continuous mappings \phi: X \to Y the graph of which is a subset of H; in other words, \kappa(H) = \{\ \phi \in E\ |\ \forall x \in X, (x, \phi(x)) \in H\ \}.

The uniform convergence topology on E is such that a subset A of E is open if and only if for any \phi \in A, there exists r > 0 such that for any \psi \in E, if \forall x \in X, \psi(x) \in \dot B(\phi(x), r), then \psi \in A. We can express this condition in terms of subsets of X \times Y that I have named open tubes: for any mapping (not necessarily continuous) \phi: X \to Y and for any r > 0, the open tube \dot T(\phi, r) is \{\ (x, y) \in X \times Y\ |\ d(\phi(x), y) < r\ \}. Then A \subseteq E is open for the uniform convergence topology if and only if for any \phi \in A, there exists r > 0 such that \kappa(\dot T(\phi, r)) \subseteq A.

The third topology ‑ the “open set” topology ‑ is the topology on E generated by all the \kappa(O) for all O open subsets of X \times Y. (The book says the topology is this collection itself, but I have not managed to prove that it directly forms a topology ‑ as is more explicitly requested in exercise 215.)

To show that the open set topology is finer than the uniform convergence one, we consider an open set for the uniform convergence topology and endeavour to show that it is open in the open set one.

Let A be an open set in the uniform convergence topology.

Let \phi be an eventual member of A.

There exists an r > 0 such that \kappa(\dot T(\phi, r)) \subseteq A.

I prove here that, \phi being continuous, \dot T(\phi, r) is an open subset of X \times Y. It includes the graph of \phi. Hence \kappa(\dot T(\phi, r)) contains \phi, and is by definition an open set for the open set topology on E. Since it is a subset of A, A is a neighbourhood of \phi for the open set topology.

Hence for any \phi \in A, A is a neighbourhood of \phi for the open set topology. This implies that A is itself an open set in the open set topology. \blacksquare

Connected subsets one of which meets the closure of the other

Exercise 224 (Chapter 32, “Connectedness”):

Let A and B be connected subsets of topological space X such that A intersects \Cl(B). Prove that A \cup B is connected. Find an example to show that it is not enough to assume that \Cl(A) intersects \Cl(B).

Let us assume that V and W are open subsets of X such that V \cap W \cap (A \cup B) is empty and that A \cup B \subseteq V \cup W. To show that A \cup B is connected, we need to show that either V or W does not intersect A \cup B.

Since A is connected, and V \cap W \cap A is empty and A \subseteq V \cup W, necessarily A is entirely in V or in W. Similarly, B is entirely in V or entirely in W.

Let us suppose that they are not both in V or both in W; for instance, that A \subseteq V and B \subseteq W.

If a \in A, then a \in V with V open in X. Since V does not intersect B, a cannot be in \Cl(B). Hence A cannot intersect \Cl(B). This contradicts our assumption.

Hence it is impossible that A \subseteq V and B \subseteq W. Similarly, it is impossible that A \subseteq W and B \subseteq V. Thus, A and B are either both in V or both in W, which implies that either V or W does not intersect A \cup B. \blacksquare

If we only assume that \Cl(A) and Cl(B) intersect, it does not always follow that A \cup B is connected. For instance, let X be the topological plane and A and B two open disks that “almost touch”, that is such that the corresponding closed disks ‑ which are their closures ‑ intersect in one point. The union of A and B is then not connected.

Union of transitively overlapping connected spaces

Exercise 225 (Chapter 32, “Connectedness”):

Prove the following generalization of theorem 40. Let A_\lambda (\lambda in \Lambda \neq \emptyset) be a collection of connected subsets of topological space X, and let the equivalence relation on this collection of sets generated by A_\lambda \approx A_{\lambda'} if A_\lambda \cap A_{\lambda'} \neq \emptyset have just one equivalence class. Then \bigcup_\lambda A_\lambda is connected.

For clarity, rather than a family (A_\lambda)_{\lambda \in \Lambda} I will consider a set \mathcal A of connected subsets of X. The condition is that the equivalence relation in \mathcal A generated by the above relation \approx has only one class.

Let K = \bigcup_{A \in \mathcal A} A.

Let V and W be eventual open subsets of X such that K \subseteq V \cup W and V \cap W \cap K = \emptyset. To show that K is connected, we must show that either V \cap K or W \cap K is empty.

First, let A be an eventual element of \mathcal A. Since A \subseteq K, the two subsets of A, A \cap V and A \cap W are disjoint, and their union is A. Furthermore, they are open in A. Since A is connected, this implies that one or the other is empty. If A \cap V is empty, then A \subseteq W; if instead A \cap W is empty, then A \subseteq V. Hence in all cases an element of \mathcal A is entirely in V or entirely in W.

Let us consider the relation \mathcal R on \mathcal A defined by A_1 \mathcal R A_2 \iff (A_1 \subseteq V \wedge A_2 \subseteq V) \vee (A_1 \subseteq W \wedge A_2 \subseteq W); in other words, A_1 \mathcal R A_2 if and only if both are subsets of V, or both are subsets of W. Clearly, this is an equivalence relation.

Furthermore, if A_1 \approx A_2, that is, A_1 \cap A_2 \neq \emptyset, we cannot have A_1 \subseteq V and A_2 \subseteq W, for V and W are disjoint. Hence A_1 and A_2 are either both in V or both in W; that is, A_1 \mathcal R A_2. Hence \mathcal R is an equivalence relation greater than the equivalence relation generated by \approx.

If \mathcal R had two or more classes, there would be some element of \mathcal A in one class and another in another; these two elements would not be equivalent following \mathcal R. Since \mathcal R is greater than the equivalence relation generated by \approx, they would not be equivalent either following this latter relation; which contradicts the fact that this has only one class. Hence \mathcal R too can have only one class, which means that for all A_1 and A_2 in \mathcal A, either both are in V or both are in W. But this implies that all are in V or all are in W, which in turn implies that their union K is a subset of either V or W, hence that eitherV \cap K or W \cap K is empty. \blacksquare

Direct products/sums: when the associated morphisms are epi/mono

Exercise 7 (Chapter 2, “Categories”):

In the category of sets, the two morphisms (\alpha and \beta) in a direct product are monomorphisms and the two morphisms in a direct sum are epimorphisms. Is this true in every category?

There are two mistakes in this wording.

  • The author has exchanged “monomorphisms” and “epimorphisms”. In a direct product of sets, the morphisms are generally not monomorphisms (injective), and in a direct sum of sets, they are generally not epimorphisms (surjective).
  • In a direct product of sets, the morphisms are almost always epimorphisms. If one of the sets is empty, the product is empty, and the morphism associated with the other set (an empty mapping) is not an epimorphism; unless this other set is empty too.

The general theorem I will prove is this:

Let A and B be objects of a category \mathfrak C, and (P, \alpha, \beta) a \mathfrak C-direct product of A and B. If \Mor(A, B) is not empty then \alpha is an epimorphism.

We consider the above conditions satisfied. There exists a morphism \gamma: A \to B.

Let \alpha' be the identity morphism A \to A and \beta' the morphism \gamma: A \to B. Applying the universal property of the direct product (P, \alpha, \beta) to the triplet (A, \alpha', \beta'), we obtain that there exists a unique \epsilon: A \to P such that both \alpha' = \alpha \circ \epsilon and \beta' = \beta \circ \epsilon.

The first of these two relations says that \alpha \circ \epsilon is the identity on A. If we have an object X and two morphisms \phi_1 and \phi_2 from A to X such that \phi_1 \circ \alpha = \phi_2 \circ \alpha, then, composing this relation on the right with \epsilon and eliminating the resulting identity morphisms, we get \phi_1 = \phi_2. Hence \alpha is an epimorphism. \blacksquare

We obtain, of course, a similar result exchanging the roles of A and B.

Reversing the arrows, we get the corresponding result for direct sums:

Let A and B be objects of a category \mathfrak C, and (S, \alpha, \beta) a \mathfrak C-direct sum of A and B. If \Mor(B, A) is not empty then \alpha is a monomorphism.

Plus, of course, the similar result obtained by exchanging A and B.

It so happens that in the category of sets, in the case of a direct product, both \alpha and \beta are always monomorphisms, even if \Mor(B, A), respectively \Mor(A, B), are empty.

Theorems about the direct product of two topological spaces

A lot of results concerning the direct product (Z, \alpha, \beta) of two topological spaces X and Y are trivial if you consider the usual particular construction of these entities. I find it more satisfying to derive them directly from the universal definitions.

See also: the equivalent page on direct sums.

In the following, we will always have X and Y topological spaces and ((Z, \tau), \alpha, \beta) a direct product thereof.

No duplicates

In the Z = X \times Y construction, this is just the trivial fact that if you know both a and b, then you know (a, b).

Theorem:

For every c_1, c_2 \in Z, if \alpha(c_1) = \alpha(c_2) and \beta(c_1) = \beta(c_2), then c_1 = c_2.

Proof:

Let us suppose that we have c_1, c_2 \in Z such that \alpha(c_1) = \alpha(c_2) and \beta(c_1) = \beta(c_2).

Let us take Z' a singleton topological space, that is that Z' is a singleton set (singleton sets exist!) and we equip it with the only possible topology, which is both the indiscrete and the discrete one.

We define \alpha' and \beta' as the constant mappings, Z' \to X and Z' \to Y respectively, that map the only element of Z' to \alpha(c_1), which is also \alpha(c_2), and to \beta(c_1), which is also \beta(c_2), respectively.

We define \gamma_1 and \gamma_2 as the constant mappings, both Z' \to Z, that map the only element of Z' to c_1 and to c_2 respectively.

Since these four mappings are constant, or, alternatively, since the topology on Z' is discrete, they are all continuous.

Furthermore, for both \gamma = \gamma_1 and \gamma = \gamma_2 we have:

    \[\alpha' = \alpha \circ \gamma\]

    \[\beta' = \alpha \circ \gamma\]

Since (Z, \alpha, \beta) is a direct product of X and Y, there can be only one morphism \gamma: Z' \to G satisfying these two relations. Hence necessarily \gamma_1 = \gamma_2, which implies c_1 = c_2. \blacksquare

Existence of an antecedent

If we have a \in X and b \in Y, does there exist a c \in Z such that \alpha(c) = a and \beta(c) = b? Again, this is trivial in the Z = X \times Y representation: c = (a, b) is the answer. Sticking to the universal definition of the direct product, we have:

For any a \in X and b \in Y, there exists c \in Z such that \alpha(c) = a and \beta(c) = b.

Proof:

We take Z' as a singleton topological space as above, and \alpha' and \beta' as the mappings Z' \to X and Z' \to Y respectively that map the unique element of Z to a and to b respectively. These two mappings again are continuous, since Z' has the discrete topology.

The universal definition of (Z, \alpha, \beta) as a direct product implies that there exists \gamma: Z' \to Z such that \alpha' = \alpha \circ \gamma and \beta' = \alpha \circ \gamma. Taking c as the image by \gamma of the unique element of Z, we thus have a = \alpha(c) and b = \beta(c). \blacksquare.

This result and the preceding one come together as:

For any a \in X and b \in Y, there exists a unique c \in Z such that \alpha(c) = a and \beta(c) = b.

The associated morphisms are surjective

At least, usually.

If X = \emptyset or Y \neq \emptyset, then \alpha is surjective. If Y = \emptyset or X \neq \emptyset, then \beta is surjective.

This is just a consequence of the general result proved in exercise 7 (corrected): in any category, if (P, \alpha, \beta) is a direct product of A and B, and if \Mor(A, B) \neq \emptyset, then \alpha is an epimorphism (and similarly, exchanging the roles of A and \alpha, and B and \beta, respectively); plus the fact that in the category of topological spaces, epimorphisms are surjective mappings.

Characterisation of the topology

That the topology on the the direct product space Z is the one generated by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y is a stipulation of the explicit construction of Z as X \times Y. It can however be proven without reference to this construction.

The topology of Z is the topology generated in Z by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y. A subset of Z is open if and only if it is the union of some collection of subsets of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open in X and V open in Y.

Proof:

Let \tau be the direct product topology on Z and \tau_0 the topology on Z generated in Z by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y.

Since \alpha and \beta are defined as continuous, necessarily \tau contains all the \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y. Hence \tau is finer than \tau_0. We wish to show that \tau_0 is also finer than \tau.

The mapping \alpha, considered from (Z, \tau_0) to X, remains continuous, since all the \alpha^\leftarrow(U) for U open in X are in \tau_0. Similarly, \beta is continuous (Z, \tau_0) \to Y. Hence we have a topological space (Z, \tau_0) and two continuous mappings \alpha and \beta, from (Z, \tau_0) to X and Y respectively. Since ((Z, \tau), \alpha, \beta) is a direct product of X and Y, there exists a unique continuous mapping \gamma: (Z, \tau_0) \to (Z, \tau) such that \alpha = \alpha \circ \gamma and \beta = \beta \circ \gamma.

This implies that any element c \in Z has the same image as \gamma(c) by both \alpha and \beta. The first theorem above implies that \gamma(c) = c. Thus \gamma = \Id_Z.

Hence \Id_Z is continous (Z, \tau_0) \to (Z, \tau); which means that \tau_0 is finer than \tau.

Hence \tau = \tau_0, that is the topology \tau of the direct product is the topology of Z generated by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y.

The open sets of the topology generated by a collection of sets \mathcal B are the unions of any number of finite intersections of elements of \mathcal B.

In this case, \mathcal B = \{ \alpha^\leftarrow(U) \}_{ U\ open\ in\ X } \cup \{ \beta^\leftarrow(V) \}_{ V\ open\ in\ Y }. A finite intersection of elements of \mathcal B will be of the form \bigcap_{i=1}^p \alpha^\leftarrow(U_i) \cap \bigcap_{j=1}^q \beta^\leftarrow(V_j) with the U_i‘s open subsets of X and the V_j‘s open subsets of Y. Since intersections are well-behaved respective to inverse images, and since finite intersections of open subsets are open subsets, this can be written \alpha^\leftarrow(\bigcap_{i=1}^p U_i) \cap \beta^\leftarrow(\bigcap_{j=1}^q V_j) = \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open subset of X and V open subset of Y. Hence an open subset of Z is a union of subsets of Z of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open subset of X and V open subset of Y\blacksquare

Continuity of a mapping to a product space

If E is a topological space and \phi a mapping E \to Z, then \phi is continuous if and only if both \alpha \circ \phi and \beta \circ \phi are continuous.

If \phi: E \to Z is continuous, since both \alpha: Z \to X and \beta: Z \to Y are continuous, by composition of continuous mappings we conclude that \alpha \circ \phi and \beta \circ \phi are continuous.

Conversely, suppose that we know that \alpha \circ \phi and \beta \circ \phi are continuous. Let W be an eventual open subset of Z. We wish to show that \phi^\leftarrow(W) is an open subset of E.

Let a be an eventual point of \phi^\leftarrow(W). Then \phi(a) \in W. Since W is open in Z and (see above) “a subset of Z is openif and only if it is the union of some collection of subsets of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open in X and V open in Y”, there exist U open in X and V open in Y such that \phi(a) \in \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) \subseteq W. Taking \phi^\leftarrow of this expression, we obtain a \in \phi^\leftarrow(\alpha^\leftarrow(U) \cap \beta^\leftarrow(V)) \subseteq \phi^\leftarrow(W). But \phi^\leftarrow(\alpha^\leftarrow(U) \cap \beta^\leftarrow(V)) = \phi^\leftarrow(\alpha^\leftarrow(U)) \cap \phi^\leftarrow(\beta^\leftarrow(V)) = (\alpha \circ \phi)^\leftarrow(U) \cap (\beta \circ \phi)^\leftarrow(V). Since we know that both \alpha \circ \phi and \beta \circ \phi are continuous, this last expression is an open subset of E, that contains a and is included in \phi^\leftarrow(W). Thus \phi^\leftarrow(W) is a neighbourhood of a. This being the case for any a \in \phi^\leftarrow(W), \phi^\leftarrow(W) is an open subset of E. \blacksquare