Let $V$ is finite dimensional vector space and $W$ is subspace of $V$. Define relation $\sim$ on $V$ in this way: for $x,y\in V$ we say that $x\sim y$ iff $x-y\in W$. Then one can show that $\sim$ is equivalence relation. Then $V/W:=V/\sim$ is the collection of all equivalence classes, i.e. $$V/W =\{x+W:x \in V\}.$$

Let's make $V/W$ a vector space. In order to do that we have to define addition and multiplication to scalars. Define these operations in the following way:

$$(x_1+W)\oplus(x_2+W):=(x_1+x_2)+W \ \text{and}\ \ \alpha\odot(x+W)=\alpha x+W$$

I want to show that operations $\oplus,\odot$ are well-defined.

To be honest I remember that I always had a trouble with the understanding of the notion well-definedness;

Basically I know that we have to show the following: If $x_1+W=y_1+W$ and $x_2+W=y_2+W$ then $(x_1+W)\oplus(x_2+W)=(y_1+W)\oplus(y_2+W)$.

But why I cannot conclude in the following way?

consider $(x_1+W)\oplus(x_2+W)$ but since $x_1+W=y_1+W, x_2+W=y_2+W$ then I can plug in and I'll get: $$(x_1+W)\oplus(x_2+W)=(y_1+W)\oplus(y_2+W)$$ but the LHS is $(x_1+x_2)+W$ and the RHS is $(y_1+y_2)+W$ and they are equal by the previous equality.

Can anyone explain from the scratch what does it mean and correct my last reasoning if its wrong? Please do not duplicate this question.


1 Answer 1


The idea of something being well-defined, is that a function does not map a point to different points in your codomain. For example, you have a map $f$ that maps $x \mapsto a$ and it maps $y \mapsto b$, you want to show that if $x=y$, then $a=b$. You cannot say that since $x=y$, it follows that $f(x)=f(y)$, because this implication only holds true when you already assume that $f$ is well-defined. Showing that $f$ is well defined means, showing that $x=y \implies f(x)=f(y)$.

Now your map $'f'$, is $\oplus : V/W \times V/W \to V/W$, which maps $(x,y) \mapsto x \oplus y $. In this case you have $$\oplus : (x_1+W , x_2 +W) \mapsto (x_1+W)\oplus(x_2 +W) $$ and you have $$\oplus : (y_1+W , y_2 +W) \mapsto (y_1+W)\oplus(y_2 +W) $$. But now, since $x_i + W = y_i + W$, you want to show that $(x_1+W)\oplus(x_2 +W) = (y_1+W)\oplus(y_2 +W) $, to prove the well-definedness of $'\oplus'$.

To prove this, you need to use that $x_i - y_i \in W$.

The story for the scalar multiplication is analogous.

  • $\begingroup$ Thanks a lot for your answer! But i still cannot get the idea. So i have to show that the "addition" $\oplus$ on $V/W$ is well defined, right? It means the following in my opinion: so i define the rule $\oplus$ on $V/W$ and then I have to show that this rule is well defined(which is the same as being function, right?). So i have to take equal pair of two cosets and show that addition will be the same. I would be grateful if you can add more details to my reasoning. $\endgroup$
    – RFZ
    Nov 13, 2019 at 18:02

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