Science Metro

Binary relation and Map

$Binary operation$

R

is a binary operation between

A

and

B

:

R \subset A \times B

$Binary operation on A$

R

is a bianry operation on

A

:

R \subset A \times A

$Map$

G_{f}

is a map from

A

to

B

:

1.\forall a \in A \exists b \in B ((a, b) \in G_{f}) 2. (a, b_{1}) \in G_{f} \land (a, b_{2}) \in G_{f} \Rightarrow b_{1} = b_{2}

f : A \to B (x \mapsto f (x)) :=

G_{f}

f (x) = y def ⟺ (x, y) \in G_{f}

f = g def ⟺ G_{f} = G_{g}

$Surjective$

f

is Surjective:

\forall b \in Y \exists a \in X (f (a) = b)

$Injective$

f

is injective:

\forall a_{1}, a_{2} \in X (f (a_{1}) = f (a_{2}) \to a_{1} = a_{2})

$Bijective$

f

is bijective

: f

is surjective

\land

f

is injective

$Identity map$

An identity map on

X

:

G_{i d_{X}} = {(x, y) \in X \times X ∣ y = x}

$Inverse map$

G_{f^{- 1}} := {(y, x) \in Y \times X ∣ f (x) = y}

$Composition operator$

Composition operator of

f, g

G_{g \circ f} := {(x, z) \in X \times Z ∣ z = g (f (x))}

$Composition map$

Composition map of

f, g

G_{g \circ f} := {(x, z) \in X \times Z ∣ z = g (f (x))}

$Corollary$

1. f, g are surjective \Rightarrow g \circ f is surjective 2. f, g are injective \Rightarrow g \circ f is injective

$Corollary$

h \circ (g \circ f) = (h \circ g) \circ f

(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1}

$Image$

X_{0}

is an image of

f

:

f (X_{0}) := {y \in Y ∣\exists x \in X_{0} (f (x) = y)}

$Corollary$

(1) X_{1} \subset X_{2} \Rightarrow f (X_{1}) \subset f (X_{2})

(2) f (X_{1} \cup X_{2}) = f (X_{1}) \cup f (X_{2})

(3) f (X_{1} \cap X_{2}) \subset f (X_{1}) \cap f (X_{2})

$Inverse image$

f

is an inverse image of

Y_{0}

:

f^{- 1} (Y_{0}) := {x \in X ∣ f (x) \in Y_{0}}

$Corollary$

(1) Y_{1} \subset Y_{2} \Rightarrow f^{- 1} (Y_{1}) \subset f^{- 1} (Y_{2})

(2) f^{- 1} (Y_{1} \cup Y_{2}) = f^{- 1} (Y_{1}) \cup f^{- 1} (Y_{2})

(3) f^{- 1} (Y_{1} \cap Y_{2}) = f^{- 1} (Y_{1}) \cap f^{- 1} (Y_{2})