The fibres of a map form a partition of the domain. This is a question from the free Harvard online abstract algebra lectures.  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see this post.
This problem is from assignment 5.

Prove that the fibres of a map form a partition of the domain.

Let $G$ and $G^\prime$ be sets and $f$ be a map from $G$ to $G^\prime$.  The nonempty fibres of $f$ correspond to the elements of the image of $f$ in $G^\prime$.  Let $R$ be a relation on $G$ defined by $a \sim b$ if $a$ and $b$ are in the same fibre.  For all $a\in G$, $f(a)=f(a)$, so $a\sim a$ and $R$ is reflexive.  Assume, for some $a,b\in G$, that $a\sim b$.  Then $f(a)=f(b)$. So $b\sim a$ and $R$ is symmetric.  Assume, for some $a,b,c\in G$, that $a\sim b$ and $b\sim c$.  Then $f(a)=f(b)$ and $f(b)=f(c)$.  So $f(a)=f(c)$, $a\sim c$, and $R$ is transitive.  Therefore, $R$ is an equivalence relation and the nonempty fibres of $f$ form a partition of $G$.
Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.
Thanks.

Your solution is correct. Since you asked for different solutions I'll give you one.
Let $f \colon G \to G'$ be a function. For each $x' \in G'$, we'll denote by $f^{-1}(\{x'\})$ the set
$$\left\{x \in G : f(x)=x' \right\}$$
i.e. the fiber of $x'$ for the map $f$.
Now given $x',y' \in G'$ then $f^{-1}(\{x'\}) \cap f^{-1}(\{y'\}) \ne \emptyset$ if and only if there exists a $z \in G$ such that $x' = f(z) = y'$ and so thus $x'=y'$.  
This implies that $f^{-1}(\{x'\}) \cap f^{-1}(\{y'\}) \ne \emptyset$ if and only if $f^{-1}(\{x'\})=f^{-1}(\{y'\})$, which means that the fibers (i.e the sets of form $f^{-1}(\{x'\})$ for some $x' \in G'$) form a partition of $G$.
