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I hope are well. I have some doubts as I am new in algebra. Let $V/W$ the quotient vector space with the usual sum and product, in addition to the equivalence relation defined on $W$ which is a subspace of $V$.

How I can prove that the canonical projection that sends a vector $v$ of $V$ to its equivalence class in $V/W$ is not injective but it's surjective?

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    $\begingroup$ It is not injective since the kernel is not zero. The kernel is $W$. $\endgroup$
    – Nick
    Apr 12, 2016 at 3:03
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    $\begingroup$ Note that it is injective if $W$ is the zero subspace $\endgroup$ Apr 12, 2016 at 3:03
  • $\begingroup$ Thank you very much for your contribution, but my biggest question is to check surjectivity, I do not see so clear =( $\endgroup$ Apr 12, 2016 at 3:06
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    $\begingroup$ Every equivalence class comes from a vector (they all look like $v+W$ for some $v \in V$), and every vector gets sent to its equivalence class by the canonical projection, hence it's subjective $\endgroup$ Apr 12, 2016 at 3:18
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    $\begingroup$ I would really wait for a different answer. I do not think universal properties are something you should be worrying about at this stage. In the meantime, pick some particular vector and quotient spaces, and see if you can understand surjectivity in some concrete cases. $\endgroup$
    – pjs36
    Apr 12, 2016 at 3:19

2 Answers 2

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I don't think Hendrik's recommendation is particularly helpful for an absolute beginner in algebra.

Here's how I'd do it:

Let $X$ denote a vectorspace and $S$ denote a subspace. We want to show that the canonical projection $X \rightarrow X/S$ is surjective.

Recall that $\mathcal{P}(X)$ is notation the collection of all subsets of $X$; this is called the powerset of $X$.

Now lets just try to remember the definitions. Let $\pi : X \rightarrow \mathcal{P}(X)$ denote the function given as follows: $$\pi(x) = x+S$$

The quotient $X/S$ is, by definition, the image of $\pi$, and the canonical projection $X \rightarrow X/S$ is, by definition, the co-restriction of $\pi$ to its image. But since every function is surjective when co-restricted to its image, hence the canonical projection is surjective.

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  • $\begingroup$ P.S. Look up words like "powerset" and "image" until this all makes complete sense. $\endgroup$ Apr 12, 2016 at 3:25
  • $\begingroup$ Thank you so much. I understood. =) $\endgroup$ Apr 12, 2016 at 3:28
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I recommend, investigate the universal property of quotient vector space.

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  • $\begingroup$ Oh, thanks, i will. $\endgroup$ Apr 12, 2016 at 3:15

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