Suppose $T \in L(V,W)$. Define $\tilde{T}: V/\ker(T) \to W$ by $\tilde{T}(\vec v + \ker(T)) = T(\vec v)$.
I am unsure as to how to show this mapping is well defined. I understand that the basic principle is to show that each X value maps to only 1 Y value but I am confused how this would work for quotient spaces as the X input.
Any help please?
On
To expand on my comment, the usual strategy when showing that maps from a quotient are well defined, is to show that it is constant on equivalence classes. Another way of phrasing this is to say that the map is "independent of representative". So, we need to take two representatives from the same equivalence class $v + \ker{T} = w + \ker{T}$ and show that $\tilde{T}(v + \ker{T}) = \tilde{T}(w + \ker{T})$.
Ok so now that we have the strategy, lets apply it. Since $v + \ker{T} = w + \ker{T}$, it then follows that $v - w \in \ker{T}$. That is, $0 = T(v-w) = T(v) - T(w) \implies T(v) = T(w)$ where the last equality follows from $T$ being a linear map.
Now, let us send $v + \ker{T}$ and $w + \ker{T}$ through the map $\tilde{T}$. $$\tilde{T}(v + \ker{T}) = T(v)$$ $$\tilde{T}(w + \ker{T}) = T(w)$$ But by the previous paragraph, we know that $T(v) = T(w)$ and we are done.
Given the edit, just taken any element $\vec k\in \ker(T)$ to get $$T(\vec v+\vec k) = T(\vec v)+T(\vec k) = T(\vec v).$$ This does not depend on our choice of $\vec k\in \ker(T)$. This is what makes it well defined; it takes the same value on the whole equivalence class $\vec v+\ker(T)$.
The point is that we apply $T$ to $\vec v+\vec k$, which is a different member of the equivalence class, we get the same value, so it makes sense to say that the value of the function $\tilde T$ on the equivalence class is equal to $T(\vec v)$, where $\vec v$ is any element in the equivalence class.