Homo and Isomorphism for Sets

126 Views Asked by At

I have three theorems about homo- and isomorphisms for sets, proving them is not really a problem but I cannot really relate them.

Theorem A Let $\sim$ be an equivalence relation on the set $A$, let $B$ be a set. Let $f : A \rightarrow B$ be a mapping such that $a_1 \sim a_2 \implies f(a_1) = f(a_2)\;$ for all $a_1, a_2 \in A$

1.) There is exactly one mapping $f' : A/\!\sim \; \rightarrow B$ with $f'([a]) \mapsto f(a)$

So far no problem, but there is more:

Definition Let $f: A \rightarrow B$ be a mapping; we define an equivalence relation on $A$ with: $a_1 \sim_f a_2 \iff f(a_1) = f(a_2)\;$ for all $a_1,a_2 \in A$

And the last one:

Theorem B Let $f : A \rightarrow B$ be a mapping, then $f':A/\!\sim_f \; \rightarrow f(A)$ is a bijection with $f'([a])= f(a)\;$ with $a\in A$

First of all $[a]$ is the equivalence class of $a$ and $A/\sim$ is the quotient set wrt. $\sim$.

So my questions:

How are these two theorems related? Is the mapping defined in A the "non-bijective" version of the mapping of Theorem B?

Is the mapping from A the same thing if it is bijective (because then $f(A) = B$ because the mapping would be surjective)? It seems so because we get the relation from the Definition if the mapping from $A$ is injective.

1

There are 1 best solutions below

0
On BEST ANSWER

Theorem B appears to be a more specific version of Theorem A, specialized to the case where the equivalence relation required by Theorem A is the one defined between the two.