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Let $ U \subseteq V $ be a subvectorspace.

Then there exists a linear mapping

$$ f: V \rightarrow V/U , v \rightarrow \bar{v} $$

where $ker(f)=U$. Why is the kernel equal to U? I know you can use the dimension formula but let's not do that right now.

The zero vector in $V/U$ is $\bar{0}= 0+U=U$.

So basically, whenever I input an element of U into the canonical projection I get a sum of vectors, which are both in U. And since U is closed under addition this is also gonna be in U. Therefore, the whole thing is in U and equal to $\bar{0}$. Is that correct?

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$$x\in \ker f\Leftrightarrow f(x)=\bar{0}\Leftrightarrow x+U=0+U\Leftrightarrow x-0\in U\Leftrightarrow x\in U.$$ That is, $\ker f=U.$

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    $x+U=0+U\Leftrightarrow x-0\in U$ Could you explain why that is? – 2017-01-28
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    It is the definition of the equivalence relation on the quotient set $V/U$: $$x\sim y\Leftrightarrow x-y\in U.$$ – 2017-01-28