Let $V$ be a vector space over a field $\mathbb{F}$ and let $W_i \subseteq\ V$ be two subspaces such that $$V=W_1+W_2$$ Let $$\gamma: V\rightarrow V/W_2$$ $$v \mapsto \bar v =v+W_2 $$ be the quotient map, which is known to be a linear transformation.
- Show that $\gamma (W_1)= V/W_2$
- Let $$\gamma_1=\gamma|w_1 : W_1\mapsto V/W_2$$ $$w_1 \mapsto \bar w_1 =w_1 +W_2$$ be the restriction of $\gamma$ . This $\gamma_1$ is known to be a linear transformation. Show that $\ker(\gamma_1) =W_1 \cap W_2$.
Any hints about how to start this question? im am at a complete lost.
Since you need help starting off, I will show half of the first question:
To show $V \diagup W_2 \subseteq \gamma (W_1)$, let $w \in V \diagup W_2$. We want to derive $w \in \gamma (W_1)$ from this. Well, to start off, $w = v + W_2$ for some $v \in V$. As $V = W_1 + W_2$, we have $v = w_1 + w_2$ for $w_i \in W_i$. So, $$w = (w_1 + w_2) + W_2 = w_1 + (w_2 + W_2) = w_1 + W_2 = \gamma (w_1) \ \text{for some } w_1 \in W_1$$ So, $w \in \gamma (W_1)$ as desired. The reverse inclusion proceeds similarly and I leave the entire second question for you.
Here is a small advice: think of quotienting of vector spaces as a kind of "$\,\text{mod}\,$" operation. For example, think of $V \diagup W_2$ almost like "$V \,\text{mod}\, W_2$". Everything in $V$ that is also in $W_2$ acts like a $0$ of $V \diagup W_2$. This is why in my answer above, $w_1 + (w_2 + W_2) = w_1 + W_2$; the $w_2$ "disappears" into $W_2$.