# Show that a linear map is well-defined

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.

• What is $\operatorname{can}(x)$? – José Carlos Santos May 5 at 9:14
• 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 $\vec{v}-\vec{w} \in U$ – Montes May 5 at 9:27

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.