Coset and Fiber

Question

The following is a remark on Universal Property of Quotients:

Let $\phi$ be a homomorphism from V to U. W $\subset$ V is a subspace such that $W\subset ker(\phi)$. Define $\phi^{-1}(u)$ as fiber over $u\in U$. Then, each coset of W is contained in a single fiber.

Could you explain why "each coset of W is contained in a single fiber?"

Answer 1

• A coset of $W$ (inside $V$) is a set of the form $v + W$, for some $v\in V$.

• If $\phi(v) = u$, then $\phi(v+v') = u$ if and only if $v'\in\ker\phi$. In other words, $\phi^{-1}(u) = v + \ker\phi$.

• If $W\subset \ker\phi$, then $v+W\subset v+\ker\phi$.

This shows that each coset is contained in some fibre. To show that it is contained in a single fibre:

• Any two fibres are either equal or disjoint. Indeed, take $\phi^{-1}(u) = u + \ker\phi$ and $\phi^{-1}(u') = u' + \ker\phi$. Now, these two are equal if and only if $u + \ker\phi = u' + \ker\phi$, or in other words $u-u'\in \ker\phi$. If they're not equal, i.e. $u-u'\not\in\ker\phi$, then $(u + \ker\phi) \cap (u' + \ker\phi)$ must be empty, because $u+k$ and $u'+k'$ have different images under $\phi$ for any $k,k'\in\ker\phi$.