Does $A\sim B$ implies $B\sim A$? In my class we defined a set $A$ to be equivalent to set $B$ (noted $A \sim B$) if and only if there exists a function $f \colon A \to B$ that is bijective. Does that imply that $B$ is equivalent to $A$ (noted $B \sim A$)?
Here is my attempt of proof:
Lets define $g \colon B \to A$
Let $b_1$ and $b_2$ be elements of $B$, then we have to show that $g(b_1)=g(b_2)$ if and only if $b_1=b_2$.
By hypothesis $g(b_1)$ and $g(b_2)$ are elements of $A$, and lets define $a_1=g(b_1)$ and $a_2=g(b_2)$.
We know that $f(a_1)=f(a_2)$ if and only if $a_1=a_2$, which means $g(b_1)=g(b_2)$.
But we know that $f(g(b_1))=b_1$, which is to say that $f(a_1)=b_1$ (This is the step I am not sure about).
So $g(b_1)=g(b_2)$ if and only if $b_1=b_2$ and so $g$ is injective.
Now about $g$ being surjective:
We want to show that for every $a_1\in A$ there exists a $b_1\in B$ such that $g(b_1)=a_1$.
By hypothesis if we have an element of $B$ called $b_1$, we know that there exists one element of $A$ called $a_1$ such that $f(a_1)=b_1$ and by substitution we get $f(g(b_1))=b_1$ and I am stucked here.
 A: Yes, if you take the inverse of $f$ you can show it's injective and surjective.
A: Yes. Let $A$ and $B$ be sets and suppose $A \sim B$, that is $A$ is related to $B$. 
We claim $B \sim A$ that is the relation $\sim$ is symmetric.
Indeed since $A \sim B$ by assumption then there exists a function $f:A \rightarrow B$ and $f$ is a bijection.
Now clearly since $f$ is a bijection then $f$ has an inverse function that is also a bijection, i.e. there exists a function $f^{-1}:B \rightarrow A$ thus $B \sim A$ and $\sim$ is symmetric.
A: You start with: "Lets define $g:B\to A$.." but there it stops. A definition of $g$ cannot be found in your effort, so actually the only thing we know about it is that it is a function $B\to A$. That makes inpossible to prove things as "$g$ is injective" or "$g$ is surjective". 
A proper way of really defining a function $g:B\to A$ is to observe that: $$\text{for every }b\in B\text{ there is a unique }a_b\in A\text{ such that }f(a_b)=b$$ Now let us prescribe function $g:B\to A$ by stating that: $$g(b)=a_b\text{ for every }b\in B$$ Then $g$ is well defined and this with $f(g(b))=f(a_b)=b$ for every $b\in B$.
Now you can go on by proving that $g$ is bijective. 
If $a\in A$ and $b:=f(a)$ then the uniqueness of $a_b$ tells us that $g(b)=a$ showing that $g$ is surjective.
If $g(b_1)=g(b_2)$ then $b_1=f(g(b_1))=f(g(b_2))=b_2$ showing that $g$ is injective.
