Function in quotient space

Question

Let $V$ and $W$ be both K-linear vector spaces. Now $Y$ is a linear subspace of $V$ and $Z$ is a linear subspace of $W$. Let $\alpha: V \rightarrow W$ be a $K$-linear function with $\alpha(Y)\subseteq Z$.

I want to prove that $\alpha$ induces linear functions $\alpha|_Y: Y \rightarrow Z$ with $\alpha|_Y(y)=\alpha(y)$ and $\bar{\alpha}: V/Y \rightarrow W/Z$ with $\bar{\alpha}(v+Y)=\alpha(v)+Z$

My idea: As $\alpha(Y)\subseteq Z$, $\exists z\in Z: \forall y\in Y:z=\alpha(y)$. So $\alpha$ induces $\alpha|_Y(y)=\alpha(y)$. $\square$

Now as $\alpha$ is linear, we can write $\bar{\alpha}(v+Y)=\bar{\alpha}(v)+\bar{\alpha}(Y)$ but now I do not know how to proceed.

Is this the right way to proof this? Thanks in advance for any kind of help.

Answer 1

Your idea is erroneously formulated: it should be $$\forall y\in Y, \exists z\in Z: z=\alpha(y),$$ which is more or less a tautology.

As to the quotient map, you only have to prove that it is well-defined, i.e. that if $y'\equiv y\mod Y$, then $\alpha(y')\equiv \alpha(y)\mod Z$, which shouldn't be very hard with the hypotheses that $\alpha$ is linear and maps $Y$ into $Z$.

Answer 2

• $\alpha(Y) \subseteq Z$ means that $\alpha(y) \in Z$ for each $y \in Y$, so, defining $\alpha|_Y : Y \to Z$ by $\alpha|_Y(y) := \alpha(y)$ for all $y \in Y$ makes sense simply because the image of an element of the domain is an element of the codomain. Note that, writting $(\exists z\in Z)(\forall y\in Y) \ z=\alpha(y)$ you are saying that every element of $Y$ goes to a same element of $Z$ under $\alpha$ (in this case, $z$) and this is not necessarily true.

• An element in the quotient $V/Y$ is an expression of the form $v+Y$, where $v \in V$. So, $v+Y$ is not the sum of $v$ with $Y$, it is the representant of $v$ in $V/Y$, the class of $v$. Two classes $v_1+Y$ and $v_2+Y$ (that is, two elements of $V/Y$) are equal whenever $v_1-v_2 \in Y$. Now, you need to prove that the rule $\overline{\alpha}(v+Y) := \alpha(v)+Z$ for every $v \in V$ is independent of the representants, in other words, no matter what class of $v$ you pick, under $\overline{\alpha}$ goes to $\alpha(v)+Z$, always. Thus, you need to prove that $$v_1+Y = v_1+Y \implies \alpha(v_1)+Z=\alpha(v_2)+Z$$ i.e. $v_1-v_2 \in Y \implies \alpha(v_1)-\alpha(v_2) \in Z$. Can you see this is true? Recall that $\alpha(Y) \subseteq Z$.