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Two Hamel bases of the same vector space have the same cardinality, so we define the dimension of a vector space as the cardinality of one of its Hamel bases. I'm trying to verify that

Let $V$ be a vector space. Let $E,F$ be subspaces of $V$ such that $E \subset F$. Then $\dim(V/F) \le \dim (V/E)$.

Could you check my below attempt?

We consider the map $$ f:V/F \to V/E, x + F \mapsto x + E. $$

Let's prove that $f$ is injective. Let $x,y\in V$ such that $f(x+F)=f(y+F)$. Then $x+E=y+E$. Then $x-y \in E$. Then $x-y \in F$ since $E \subset F$. Hence $x+F=y+F$. It's clear that $f$ is a homomorphism. The claim then follows.

Analyst
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    If this map was well defined, it would also be surjective as well, right? Since $x + E$ will be mapped to by $x + F$. As such I don't think this map is well-defined, and I would suggest a similar map in the opposite direction. – Daniel May 21 '23 at 21:34
  • You would also need to show that if $x+F=y+F$, then $x+E=y+E$, to show the map is well-defined, and you are going to have a bit of a problem proving that... I mean, consider the case of $E={\mathbf{0}}$ and $V=F$. Your definition does not yield a well-defined function, let alone and injection. – Arturo Magidin May 21 '23 at 21:38
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    Instead, consider the map going the other way, and use Rank-Nullity. – Arturo Magidin May 21 '23 at 21:39
  • No, the correct conclusion is $\dim(V/F)\leq\dim(V/E)$. – Arturo Magidin May 21 '23 at 21:56
  • @ArturoMagidin It's my bad... Thank you so much for your and Daniel's help! – Analyst May 21 '23 at 21:57
  • @ArturoMagidin I have a recent question about Fredholm alternative. If you don't mind, please have a look at it. Thank you so much! – Analyst May 22 '23 at 08:47

1 Answers1

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As Arturo and Daniel pointed out in their comment, the correct map should be $$ f:V/E \to V/F, x + E \mapsto x + F. $$

Let's prove that $f$ is well-defined. Let $x,y\in E$ such that $x+E=y+E$. Then $x-y\in E$. Then $x-y\in F$ since $E \subset F$. Then $x+F=y+F$. Then $f(x+E)=f(y+E)$. Clearly, $f$ is linear and surjective. By rank-nullity theorem, we get $\dim (V/E) = \dim (V/F) + \dim (\ker f)$. The claim then follows.

Analyst
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