Section 3: Problem 4 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Let $f:A\rightarrowВ$ be a surjective function. Let us define a relation on $A$ by setting $a_{0}\sim a_{1}$ if $f(a_{0})=f(a_{1}).$

(a) Show that this is an equivalence relation.

(b) Let $A^{*}$ be the set of equivalence classes. Show there is a bijective correspondence of $A^{*}$ with $B$ .

(a) RST clearly hold (they are inherited from the corresponding properties of the $=$ relation: $f(a)=f(a)$ , $f(a)=f(b)$ $\Rightarrow f(b)=f(a)$ , and $f(a)=f(b)=f(c)$ $\Rightarrow f(a)=f(c)$ ).

(b) Let $g:A^{*}\rightarrow B$ be a function such that $g(a^{*})=f(a)$ where $a\in a^{*}$ . $g$ is well-defined, as the choice of $a\in a^{*}$ does not matter, injective, as $f$ is different for elements in different classes (otherwise they would be in the same class), and surjective, as $f$ is surjective.

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Section 3: Problem 4 Solution

Section 3: Problem 4 Solution