This is a question from the free Harvard online abstract algebra [lectures](http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra). I'm posting my solutions here to get some feedback on them. For a fuller explanation, see [this](https://math.stackexchange.com/questions/98291/autv-is-isomorphic-to-s-3) post.

This problem is from assignment 5.

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