Solution:

• We need to show that $\tilde{T} $ is well defined, that is, if $v_1 + \ker T = v_2 + \ker T,~ v_1, v_2 \in V$, then the image should be same. Note that the above implies \[ v_1 - v_2 \in \ker T \implies T\left( v_1 - v_2\right) = 0 \implies T\left( v_1 \right) = T\left( v_2 \right) . \] Now observe that \begin{align*} \tilde{T}\left( v_1 + \ker T \right) = T\left( v_1 \right) = T\left( v_2 \right) = \tilde{T} \left( v_2 + \ker T \right). \end{align*} Hence, $\tilde{T}$ is well defined.

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• Let $v_1, v_2 \in V$ and $\alpha \in \mathbb{R} $. We need to show that \[ \tilde{T} \left( \alpha \cdot (v_1 + \ker T) + (v_2 + \ker T)\right) = \alpha \cdot \tilde{T} \left( v_1 + \ker T \right) + \tilde{T} \left( v_2 + \ker T \right). \] Recall that that \[ \alpha \cdot (v_1 + \ker T) + (v_2 + \ker T) = (\alpha v_1 + v_2) + \ker T. \] Therefore, \begin{align*} \tilde{T} \left( \alpha \cdot (v_1 + \ker T) + (v_2 + \ker T)\right) & = \tilde{T} \left( \alpha v_1 + v_2 + \ker T \right) \\ & = T \left( \alpha v_1 + v_2 \right) \\ & = \alpha T\left( v_1 \right) + T\left( v_2 \right), \end{align*} as $T$ is linear.

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• We need to show that $\tilde{T} $ is injective. Let \begin{align*} \tilde{T} \left( v_1 + \ker T \right) = \tilde{T} \left( v_2 + \ker T \right) & \implies T\left( v_1 \right) = T\left( v_2 \right) \\ & \implies T\left( v_1 - v_2 \right) = 0 \\ & \implies v_1 - v_2 \in \ker T \\ & \implies v_1 + \ker T = v_2 + \ker T. \end{align*}

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• We need to show that $\operatorname{im}\left( \tilde{T} \right) = \operatorname{im}\left( T \right) $. Let $w \in \operatorname{im}\left( \tilde{T} \right) $, this implies there exists $v\in V$ such that \[ w = \tilde{T} \left( v + \ker T \right) = T(v). \] Therefore, $w \in \operatorname{im}\left( T \right) $ and hence, \[ \operatorname{im}\left( \tilde{T} \right) \subseteq \operatorname{im}\left( T \right) . \] On the other hand, take $w\in \operatorname{im}\left( T \right) $. So, there exists $v\in V$ such that $T(v) = w$. Since $v\in V$ so, $v + \ker T \in V/\ker T$. Now, \[ \tilde{T} \left( v + \ker T \right) = T(v) = w, \] which implies $w \in \operatorname{im}\left( \tilde{T} \right) $ and hence \[ \operatorname{im}\left( T \right) \subseteq \operatorname{im}\left( \tilde{T} \right) . \] Therefore, we have the equality.

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• Finally, we need to show that $V/\ker T$ is isomorphic to $\operatorname{im}\left( T \right) $. Note that $\tilde{T} $ is an isomorphism between these two spaces.