VIII. Let p be an equivalence on a set X. A subset A ⊂ X is said to be a transversal of p if eac

Step 1: Let $a, b in H_{A,p}$. Then $im(a) = im(b) = A$ and $ker(a) = ker(b) = p$.

Question

VIII. Let $p$ be an equivalence on a set $X$. A subset $A \subset X$ is said to be a transversal of $p$ if each $p$-class contains exactly one element of $A$. In the full transformation semigroup $T_X$, denote the $H$-class $H_{A, p} = \{a \in T_X : im(a) = A, ker(a) = p\}$ Show that $H_{A, p}$ is a group if and only if $A$ is a transversal of $p$. Hint: you may use Green's theorem.

          VIII. Let $p$ be an equivalence on a set $X$. A subset $A \subset X$ is said to be a transversal of $p$ if each $p$-class contains exactly one element of $A$.
In the full transformation semigroup $T_X$, denote the $H$-class
$H_{A, p} = \{a \in T_X : im(a) = A, ker(a) = p\}$
Show that $H_{A, p}$ is a group if and only if $A$ is a transversal of $p$.
Hint: you may use Green's theorem.

VIII. Let p be an equivalence on a set X. A subset A ⊂ X is said to be a transversal of p if each p-class contains exactly one element of A.
In the full transformation semigroup TX, denote the H-class
HA, p = {a ∈ TX : im(a) = A, ker(a) = p}
Show that HA, p is a group if and only if A is a transversal of p.
Hint: you may use Green's theorem.

Added by Sarah F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

03/12/2023

Step 1

Then $im(a) = im(b) = A$ and $ker(a) = ker(b) = p$. Show more…

Show all steps

Thanks for your feedback!

VIII. Let p be an equivalence on a set X. A subset AC X is said to be a transversal of p if each p-class contains exactly one element of A. In the full transformation semigroup Tx, denote the H-class $H_{A,p} = \{a \in T_X : im(a) = A, ker(a) = p\}$ Show that $H_{A,p}$ is a group if and only if A is a transversal of p. Hint: you may use Green's theorem.