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Given a set $X$ and an equivalence relation $\sim$ on $X$, define quotient mapping $\pi:X \to X/\sim$, it's known the canonical projection is surjective and this fact follows from definition of equivalence class and the symmetric property of $\sim$.

Theorem:

Define $f:\small X \to B$ such that for every $a,b \in X$ if $a \sim b$ implies $f(a)=f(b)$, then there exist a unique function $G:\small {X/\sim} \to B$, such that $f = g\pi$. If f is a surjection and $a \sim b ↔ f(a) = f(b)$, then $g$ is a bijection.

If $f$ is surjective then so is $g$, form where the injectively of $g$ follows?and why $g$ is unique?

Also what is the application of this theorem?

1 Answers1

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$g$ is unique because it is determined by $[x]\mapsto f(x)$ where $[x]$ denotes the equivalence class represented by $x$.

If $h$ would do the same then $h([x])=f(x)=g([x])$ for every $x$ hence $h=g$.

$g$ is injective because $g([x])=g([y])$ implies that $f(x)=g([x])=g([y])=f(y)$ or equivalently $x\sim y$ or equivalently $[x]=[y]$.

Application of the theorem...

Well, it appears that by surjective $f:X\to B$ there is no essential difference between $X/\sim$ (with elements $[x]$) and $B$ (with elements $f(x)$) and no essential difference between $f$ and $\pi$. We can mark $f$ as a quotient function, just like the canonical $\pi$. It is a good thing to have a bright view on that.

drhab
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  • You say $[x]\mapsto f(x)$, clearly by definition of function ( left-totality) for every $[x]$ in $X/\sim$ there exist a $g([x])$ such that $[x]\mapsto g([x])$, but how do you know $g([x])=f(x)$? –  Feb 01 '20 at 12:58
  • $g([x])=f(x)$ for every $x\in X$ is exactly the statement that $g\circ\pi=f$. – drhab Feb 01 '20 at 13:00
  • Also you say $f(x)=f(y)$ or equivalently $x \sim y $ but by definition of $f$ we have $a \sim b$ implies $f(a)=f(b)$ and not vice versa, so how..? –  Feb 01 '20 at 13:00
  • I read in your question: "...If $f$ is a surjection and $a\sim b\iff f(a)=f(b)$". So two sides. Btw, if it would not be two sides then how is $\sim$ defined?... – drhab Feb 01 '20 at 13:02
  • The link I've seen the theorem is https://en.wikipedia.org/wiki/Equivalence_relation , do you think the direction should holds for the other side? (if yes so there is an error in Wikipedia) –  Feb 01 '20 at 13:04
  • If we only have $a\sim b\implies f(a)=f(b)$ then the proof in my answer concerning uniqueness of $g$ is still valid. It rests completely on $g([x])=f(x)$ and I do not use the other side there. For injectivity we need more (this is also clear on Wikipedia). – drhab Feb 01 '20 at 13:12
  • so how injectively of $g$ can be proven? –  Feb 01 '20 at 13:13
  • Injectivity requests the other side. Nowhere in your question it is asserted that there is also injectivity if the other side lacks. Actually it is about uniqueness of $g$ and secondly bijectivity of $g$ and the condition given for bijectivity are: "$f$ is surjective and $a\sim b\iff f(a)=f(b)$." – drhab Feb 01 '20 at 13:21
  • For instance let it be that $a\sim b\iff a=b$. Then $X/\sim={{x}\mid x\in X}$ and $\pi:X\to X/\sim$ is prescribed by $x\mapsto[x]={x}$. Evidently $a\sim b\implies f(a)=f(b)$ and $g$ is the function prescribed by $g({x})=f(x)$. If in that situation $f$ is not injective then also $g$ is not injective. This shows the necessity of the other side for $g$ being injective. – drhab Feb 01 '20 at 13:36
  • @Thank you, I don't have the book that this question comes from and it appears that Wiki is wrong. can you do me a favor and look at this post of mine? https://math.stackexchange.com/questions/3525445/on-the-fundamental-theorem-of-equivalence-relations –  Feb 01 '20 at 14:37
  • Is there still something that withholds you from accepting my answer? Please let me know. – drhab Feb 02 '20 at 11:47