Map composition

-

We can define a new map by composing two or more maps. Map composition can be regarded as an operation between maps. As such, we can make an algebra of maps.



Definition (map composition)

Let \(f: A \to B\) and \(g: B \to C\) be maps. In particular, note that \(\text{cod}f = \text{dom}g\). Then, we define a new map, denoted \(g\circ f: A \to C\), by

\[(g\circ f)(x) = g(f(x))\]

for each \(x \in A\). This new map \(g\circ f\) is called the composition of \(f\) with \(g\). It is convenient to read \(g\circ f\) as "\(g\) after \(f\)."

The composition can be graphically represented as in the following commutative diagram:


Remark. As noted in the definition, we can compose \(g\circ f\) if and only if \(\text{cod}f = \text{dom}g\).
Remark. Some authors (e.g., in computer science) prefer to write \(f; g\) ("\(f\), then \(g\)") rather than \(g\circ f\).
Example. Suppose that
\[f: \mathbb{Z} \to \mathbb{Z}; f(x) = x - 1,\]
and
\[g: \mathbb{Z} \to \mathbb{N}_{0}; g(x) = x^2.\]
Then, \[g\circ f: \mathbb{Z} \to \mathbb{N}_{0}; (g\circ f)(x) = x^2 - 2x + 1.\]
(Recall that \(\mathbb{N}_0 = \mathbb{N}\cup\{0\}\).)

Theorem

Let \(f: A \to B\) and \(g: B \to C\) be maps. 
  1. If \(f\) and \(g\) are injective, then \(g\circ f\) is injective.
  2. If \(f\) and \(g\) are surjective, then \(g\circ f\) is surjective.
Proof.
  1. We assume that \(f\) and \(g\) are injective. Suppose \(a, a'\in A\) and \(a \neq a'\). Since \(f\) is injective, we have \(f(a) \neq f(a')\). Since \(g\) is injective, \(g(f(a)) \neq g(f(a'))\). That is, \((g\circ f)(a) \neq (g\circ f)(a')\). Thus, \(g\circ f\) is injective.
  2. We assume that \(f\) and \(g\) are surjective. Since \(g\) is surjective, for any \(c \in C\), there exists some \(b \in B\) such that \(c = g(b)\). For this \(b\), since \(f\) is surjective, there exists some \(a \in A\) such that \(b = f(a)\). In summary, for any \(c \in C\), there exists some \(a\in A\) such that \(c = g(f(a))\), that is, \((g\circ f)(a) = c\). Hence, \(g\circ f\) is surjective.
■

Corollary

The composition of two bijections is a bijection.

Theorem

Let \(f: A \to B\), \(g: B\to C\), and \(h: C \to D\) be maps. Then, we have
\[(h\circ g)\circ f = h\circ(g\circ f).\]
In other words, map composition is associative. This can be graphically represented in the following diagram:
Proof. For any \(x \in A\), we have
\[\begin{eqnarray} ((h\circ g)\circ f)(x) &=& (h\circ g)(f(x))\\ &=&h(g(f(x))\\ &=&h((g\circ f)(x))\\ &=& (h\circ (g\circ f))(x). \end{eqnarray}\] 
■
Because of this associativity, we can denote both \((h\circ g)\circ f\) and \(h\circ(g\circ f)\) as \(h \circ g \circ h\) without parentheses.

Theorem

Let \(\alpha: X \to Y\) be a map. The following are equivalent
  1. The map \(\alpha\) is a bijection.
  2. There exists a map \(\beta: Y \to X\) such that \(\beta \circ \alpha = \text{Id}_{X}\) and \(\alpha \circ \beta = \text{Id}_{Y}\).
Proof. (\(1 \implies 2\)) Suppose \(\alpha\) is a bijection. This means that for each \(y \in Y\), there exists a unique \(x\in X\) such that \(y = \alpha(x)\). Using this unique \(x\), let us define \(\beta(y) = x\). Thus, for each \(y \in Y\), \(\beta(y) \in X\) is uniquely determined, and hence \(\beta: Y \to X\) is a well-defined map. By this definition, for each \(x \in X\), we have \(\beta(\alpha(x)) = x\). Thus, \(\beta\circ \alpha = \text{Id}_X.\) Similarly, for each \(y \in Y\), we have \(\alpha(\beta(y)) = y\), and hence \(\alpha\circ \beta = \text{Id}_Y.\)

(\(2 \implies 1\)) Suppose that such \(\beta\) as given above exists. Then, for each \(y \in Y\), \((\alpha\circ \beta)(y) = \alpha(\beta(y))= \text{Id}_Y(y) = y\) so that \(y \in \text{im}\alpha\). This means that \(Y \subset \text{im}\alpha\). \(\text{im}\alpha \subset Y\) is trivial. Hence, \(Y = \text{im}\alpha\), and \(\alpha\) is surjective. Next, suppose that \(x, x'\in X\) and \(\alpha(x) = \alpha(x')\). Apply \(\beta\) on both sides, and we have \(\beta(\alpha(x)) = \beta(\alpha(x'))\). But, \(\beta\circ\alpha = \text{Id}_X\) so we have \(\text{Id}_X(x) = \text{Id}_X(x')\), and hence \(x = x'\). Thus, \(\alpha\) is injective. In summary, \(\alpha\) is both surjective and injective, so it is bijective.
■
Remark. The map \(\beta\) in the above theorem is called the inverse map of \(\alpha\) and is denoted as \(\alpha^{-1}\). It should be obvious that the inverse map of \(\beta = \alpha^{-1}\) is \(\alpha\). □



Comments

Post a Comment

Popular posts from this blog

Open sets and closed sets in \(\mathbb{R}^n\)

-
Image
Open sets In \(\mathbb{R}\), we have the notion of an open interval such as \((a, b) = \{x \in \mathbb{R} | a < x < b\}\). We want to extend this idea to apply to \(\mathbb{R}^n\). We also introduce the notions of bounded sets and closed sets in \(\mathbb{R}^n\). Recall that the \(\varepsilon\)-neighbor of a point \(x\in\mathbb{R}^n\) is defined as \(N_{\varepsilon}(x) = \{y \in \mathbb{R}^n | d(x, y) < \varepsilon \}\) where \(d(x,y)\) is the distance between \(x\) and \(y\). Definition (Open set) A subset \(U\) of \(\mathbb{R}^n\) is said to be an  open set  if the following holds: \[\forall x \in U ~ \exists \delta > 0 ~ (N_{\delta}(x) \subset U).\tag{Eq:OpenSet}\] That is, for every point in an open set \(U\), we can always find an open ball centered at that point, that is included in \(U\). See the following figure. Perhaps, it is instructive to see what is  not  an open set. Negating (Eq:OpenSet), we have \[\exists x \in U ~ \forall \delta > 0 ~ (N_{\delta}(x) \not
Read more

Euclidean spaces

-
Image
We would like to study multivariate functions (i.e., functions of many variables), continuous multivariate functions in particular. To define continuity, we need a measure of "closeness" between points. One measure of closeness is the Euclidean distance. The set \(\mathbb{R}^n\) (with \(n \in \mathbb{N}\)) with the Euclidean distance function is called a Euclidean space. This is the space where our functions of interest live. The real line is a geometric representation of \(\mathbb{R}\), the set of all real numbers. That is, each \(a \in \mathbb{R}\) is represented as the point \(a\) on the real line. The coordinate plane , or the \(x\)-\(y\) plane , is a geometric representation of \(\mathbb{R}^2\), the set of all pairs of real numbers. Each pair of real numbers \((a, b)\) is visualized as the point \((a, b)\) in the plane. Remark . Recall that \(\mathbb{R}^2 = \mathbb{R}\times\mathbb{R} = \{(x, y) | x, y \in \mathbb{R}\}\) is the Cartesian product of \(\mathbb{R}\) with i
Read more

Applications of multiple integrals

-
Image
We can use multiple integrals to compute areas and volumes of various shapes. Area of a planar region Definition (Area) Let \(D\) be a bounded closed region in \(\mathbb{R}^2\). \(D\) is said to have an area if the multiple integral of the constant function 1 over \(D\), \(\iint_Ddxdy\), exists. Its value is denoted by \(\mu(D)\): \[\mu(D) = \iint_Ddxdy.\] Example . Let us calculate the area of the disk \(D = \{(x,y)\mid x^2 + y^2 \leq a^2\}\). Using the polar coordinates, \(x = r\cos\theta, y = r\sin\theta\), \(dxdy = rdrd\theta\), and the ranges of \(r\) and \(\theta\) are \([0, a]\) and \([0, 2\pi]\), respectively. Thus, \[\begin{eqnarray*} \mu(D) &=& \iint_Ddxdy\\ &=&\int_0^a\left(\int_0^{2\pi}rd\theta\right)dr\\ &=&2\pi\int_0^a rdr\\ &=&2\pi\left[\frac{r^2}{2}\right]_0^a = \pi a^2. \end{eqnarray*}\] □ Volume of a solid figure Definition (Volume) Let \(V\) be a solid figure in the \((x,y,z)\) space \(\mathbb{R}^3\). \(V\) is sai
Read more