Show that the map $ Γ : \mathrm{Hom}(V /W, V') → \mathrm{Hom}(V, V ')$ given by $Γ(T) =T ◦ π $ is injective

Question

Let $W$ be a subspace of $V$. Define a map $Γ : \mathrm{Hom}(V /W, V') → \mathrm{Hom}(V, V ')$ by $ Γ(T) =T ◦ π$, where π is the natural quotient map from $V → V /W$, and $T ∈ \mathrm{Hom}(V /W, V ')$.

Show that Γ is an injective linear map. Find the range of Γ.

i know that we need to show that kernel of Γ equal= {0} .

so i started with Γ (T)=0 which means $T ◦ π$=$0$ i cant seem to proceed from here ,where $V ,W$ and $V'$ are vector spaces

Answer 1

• If $T \in \ker \Gamma$, then $T\circ\pi=0$, and then for all $v \in V$ we have $$0 = 0(v) = (T\circ\pi)(v) = T(\pi(v)),$$ but since every $z \in V/W$ is $\pi(v)$ for some $v \in V$, it follows that $T(z) = 0$ for all $z \in V/W$, that is, $T = 0$. Hence $\ker \Gamma = \{0\}$.
• Take $f \in \mathcal{L}(V,V')$ such that $W \subseteq \ker f$. Then, for $z \in V/W$ define $T(z) := f(v)$ in the case that $z = \pi(v)$ for some $v \in V$. Now observe that if $z \in V/W$ can be written in two ways as $z = \pi(v_1)$ and $z = \pi(v_2)$ for some $v_1,v_2 \in V$, it follows that $v_1-v_2 \in W$, and then $v_1-v_2 \in \ker f$, which means $$0 = f(v_1-v_2) = f(v_1)-f(v_2).$$ Therefore, $T(z)$ is well-defined for all $z \in V/W$, and hence $T : V/W \to V'$. Finally, it is easy to show that $T \in\mathcal{L}(V/W,V')$ and that $T \circ \pi = f$. Hence $$\operatorname{range}\Gamma = \{f \in \mathcal{L}(V,V') : W \subseteq \ker f\}.$$

Answer 2

Assume $V, W, V'$ are vector spaces. Take $T : V/W \rightarrow V'$ a linear map. Let $\pi : V \rightarrow V/W$ be the canonical map. Define $\Gamma(T) = T \circ \pi : V \rightarrow V'$. The goal is to show that $\ker(\Gamma) = 0$. Suppose $T \circ \pi = 0$. Notice that $\pi$ is surjective, so for all $x + W \in V/W$ we have $x \in V$ with $\pi(x) = x + W$. Notice that $T \circ \pi = 0$ implies that for all $x \in V$, $T \circ \pi(x) = T(x + W) = 0$, but by surjectivity this means that $T = 0$ (since it evaluates to $0$ on every element). Thus $\ker(\Gamma) = 0$.