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Let $V,W$ be vector spaces over field $F$ and $f:V\rightarrow W$ be a linear mapping such that $U \subseteq \ker(f)$. Show that the mapping $\overline{f}:V/U \rightarrow W$ given by $\overline{f}(v+U)=f(v)$ is a well-defined linear mapping. Show that $\ker(\overline{f})$=can$(\ker(f))$ where can$(\ker(f))=\{$can$(x) \mid x \in \ker(f)\}$

It looks like I need to apply first isomorphism theorem but I have no idea how to do that.

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    $\begingroup$ What is $\operatorname{can}(x)$? $\endgroup$ – José Carlos Santos May 5 at 9:14
  • $\begingroup$ It is a canonical map defined by can$:V\rightarrow V/U$ such that $\vec{v} \mapsto E(\vec{v})$ where $E(\vec{v})$ are equivalence classes of the vectors $\vec{v}$ under the equivalence relation $\vec{v} \sim \vec{w}$ if $\vec{v}-\vec{w} \in U$ $\endgroup$ – Montes May 5 at 9:27
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For your first question: if $v+U = w+U$, then $v-w\in U\subseteq \mathrm{ker}(f)$ and so $f(v) = f(w)$. Hence $\bar{f}(v+U) = f(v) = f(w) = \bar{f}(w+U)$. Thus, $\bar{f}$ is well-defined.

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