surjective map

[logistic_guy]

Let $\displaystyle f \ : \ A \rightarrow B$ be a surjective map of sets. Prove that the relation $\displaystyle a \sim b$ if and only if $\displaystyle f(a) = f(b)$ is an equivalence relation whose equivalence classes are the fibers of $\displaystyle f$.

 

[khansaheb]

Let $\displaystyle f \ : \ A \rightarrow B$ be a surjective map of sets. Prove that the relation $\displaystyle a \sim b$ if and only if $\displaystyle f(a) = f(b)$ is an equivalence relation whose equivalence classes are the fibers of $\displaystyle f$.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

To prove that the relation is an equivalence relation, I must prove or show that:

$\displaystyle \sim$ is reflexive.

And

$\displaystyle \sim$ is symmetric.

And

$\displaystyle \sim$ is transitive.

 

[logistic_guy]

If the preimage of the element $\displaystyle a \in A$ under the map $\displaystyle f$ is:

$\displaystyle f^{-1}(a) = \{x \in A \ | \ x = a\}$

Then,

The preimage of the element $\displaystyle b \in B$ under the map $\displaystyle f$ is:

$\displaystyle f^{-1}(b) = \{x \in A \ | \ f(x) = b\}$

 

[pka]

Let $\displaystyle f \ : \ A \rightarrow B$ be a surjective map of sets. Prove that the relation $\displaystyle a \sim b$ if and only if $\displaystyle f(a) = f(b)$ is an equivalence relation whose equivalence classes are the fibers of $\displaystyle f$.

If $P \& Q$ are sets and $f:P \to Q$ surjectionly then $\overleftarrow{f}(t)$ is the set $\{x\in P: f(x)=t\}$
Because $f$ is a surjection, we know $\overleftarrow{f}(t)\not=\emptyset$ for all $t\in Q$ WHY IS THAT TRUE?
You seem to under that $\forall {a\in P \&\ b\in P}$ then $(a\sim b)\iff(f(a)=f(b))$
Do you have any issues with the relation $\sim$ being an equivalence relation?
Do you see that $(\forall t\in Q)$ the collection of sets $\overleftarrow{f}(t)$ forms a partition of set $P$?
Now as fibers I have a huge collection of textbooks in set theory &/or function theory none of which use that term.
But I do suspect that the sets $\overleftarrow{f}(t)$ are what they mean.

 

[logistic_guy]

If $P \& Q$ are sets and $f:P \to Q$ surjectionly then $\overleftarrow{f}(t)$ is the set $\{x\in P: f(x)=t\}$
Because $f$ is a surjection, we know $\overleftarrow{f}(t)\not=\emptyset$ for all $t\in Q$ WHY IS THAT TRUE?
You seem to under that $\forall {a\in P \&\ b\in P}$ then $(a\sim b)\iff(f(a)=f(b))$
Do you have any issues with the relation $\sim$ being an equivalence relation?
Do you see that $(\forall t\in Q)$ the collection of sets $\overleftarrow{f}(t)$ forms a partition of set $P$?
Now as fibers I have a huge collection of textbooks in set theory &/or function theory none of which use that term.
But I do suspect that the sets $\overleftarrow{f}(t)$ are what they mean.

Yeah the term fiber (fibers) looks weirdo, but it is widely used in Russia and the Soviet Union in general.

For example, the fiber of $\displaystyle f$ over $\displaystyle f(a)$ is just $\displaystyle f^{-1}(f(a))$. Other notations can be used and yours $\left(\overleftarrow{f}(f(a))\right)$ may be one of them.

In other words, it is the same as saying $\displaystyle f^{-1}(f(a))$ is the preimage of the element $\displaystyle f(a)$ under the function $\displaystyle f$. Of course it consists of all the points in the domain $\displaystyle A$ that are mapped to the same value $\displaystyle f(a)$ in the codomain $\displaystyle B$.

Thanks a lot professor pka for passing by. What you have given is very valuable and I'll study it deeply.
Be in touch with this thread if you wanna see how I'll crack the h-ell of this elementary problem.

 

[logistic_guy]

Let us check the first property which is reflexivity.

To show $\displaystyle \sim$ is reflexive means we need to show $\displaystyle a \sim a$. We know that $\displaystyle f(a) = f(a)$, then it is absolutely true that $\displaystyle a \sim a$ and $\displaystyle \sim$ is reflexive.

 

[logistic_guy]

Let us check the second property which is symmetry.

To show $\displaystyle \sim$ is symmetric means we need to show if $\displaystyle a \sim b$, then $\displaystyle b \sim a$. We know that $\displaystyle f(a) = f(b)$, then $\displaystyle f(b) = f(a)$. This means that $\displaystyle b \sim a$ and $\displaystyle \sim$ is symmetric.

 

[logistic_guy]

Let us check the third property which is transitivity.

To show $\displaystyle \sim$ is transitive means we need to show if $\displaystyle a \sim b$ and $\displaystyle b \sim c$, then $\displaystyle a \sim c$. We know that $\displaystyle f(a) = f(b)$ and $\displaystyle f(b) = f(c)$, then $\displaystyle f(a) = f(c)$ and $\displaystyle \sim$ is transitive.

 

[logistic_guy]

Now an equivalent class of that relation is just the set $\displaystyle f^{-1}(a) = \{x \in A \ | \ x = a\}$ and the fiber of $\displaystyle f$ of any element $\displaystyle b \in B$ is just the set $\displaystyle f^{-1}(b) = \{x \in A \ | \ f(x) = b\}$.

Since the the function $\displaystyle f$ is surjective, then every element in $\displaystyle B$ has the form $\displaystyle f(a)$ for any $\displaystyle a \in A$.

Then,

The fibers of the function $\displaystyle f$ are the equivalence classes of that relation.