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I'm trying to under the definition of an embedding mapping $\pi:X \to X/V$ where $V\subset X$ are linear spaces. I have the definition $\pi (x) = x+V$ where $x+V = \{y\in X: y-x\in V \}$.

Why is $V\subset \ker \pi$?

$\ker \pi$ is the set $\{ a\in X \;st\;\forall y\in X : y-a\notin V\}$

So then if $v\in V \implies y-v\notin V\;\forall y\in X$

But what about $v$ itself? $v-v=0\in V$

What am I missing?

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    π is not an embedding, and V is not contained in ker π: it *is* ker π. – 2017-02-09
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    Here the linear mapping $\pi:X\to X/V$ would commonly be called a *quotient map*, and $X/V$ called a *quotient space*. – 2017-02-10

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Your very first sentence:

$\ker \pi$ is the set $\{ a\in X \;st\;\forall y\in X : y-a\notin V\}$

is incorrect. In fact $\ker \pi$ is the set of $a \in X$ with $\pi(a)$ being the additive identity element in the target space, $X/V$. That identity element is the set $V$. Thus $\ker \pi$ is the set of $a$ such that $a + V = V$. Something is in $a+V$ exactly if it's $a$ plus something in $V$. Well: $a$ is $a + 0$, and $0 \in V$. Hence $a \in V$. So all elements of $\ker \pi$ are in $V$. And (evidently) all elements of $V$ are in $\ker \pi$. So $\ker \pi = V$.

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    Still not understanding cause am not really used to your math speak. More used to working with sets. Is it correct to say that $x+V$ is the set of linear translations on $x$ that end up in $V$? Then $ker \pi$ would be the set of elements in $X$ that can never be linearly translated to inside $V$ except when using the zero vector. And that includes all of $V$? But how is it necessary for an element in $V$ to end up outside of $V$ if translated by a non-zero vector? – 2017-02-09
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    "Math speak" is how we talk about math. Complaining about it is like going to the zoo and asking why there are all those animals. To answer your "Is it correct to say...?" question: no. That's not a correct statement. Indeed, "linear translations" doesn't seem (to me) to have a well-defined meaning, so the statement isn't really even meaningful (to me). Translations are a certain kind of function; the only linear one (on a vector space) is the identity. – 2017-02-09