Vector spaces-First isomorphism theorem

Let $T:V \rightarrow W$ be a linear transformation between vector spaces $V$ and $W$.

Then, $\tau:T/ker(W) \rightarrow Im(W)$ induces an isomorphism given by $\tau(v+ker(T)) = T(v)$.

First we need to establish that if $v+ket(T)$ is replaced by $v'+ker(T)$ for $v,v'$ in the same coset ie $v-v' \in ker(T)$, then  $T(v)=T(v')$. That is, we need to establish that above map is well defined.

Notice,

$T(v) = T( (v'-v)+v' ) = T(v-v')+T(v') = T(v')$

Then we need to establish that the map $\tau$ is a linear map.

That is, for $v,v' \in V$, we need to show that $\tau(v+ker(T) + v'+ker(T))=\tau(v+ker(T))+\tau(v'+ker(T))$.

Indeed,  $\tau(v+ker(T) + v'+ker(T)) = T(v+v'+ker(T)) =T(v)+T(v')=\tau(v+ker(T))+\tau(v'+ker(T)$.

And

$\tau(\alpha (v+ker(T))) = \tau(\alpha v + \alpha  ker(T) ) = T(\alpha v) = \alpha T(v) = \alpha \tau(v+ker(T))$.

Thus the map T/ker(W)$ is a linear map.

To prove isomorphism, we need to show that this map is one-one and onto.

For one-one, if $\tau(v+ker(T))=0$, need to show that $v+ker(T)=0$ which is a direct result of the observation that if $\tau(v+ker(T))=T(v)=0$, then $v \in ker(T)$.

For onto, just note any element of $im(T)$ can be written as $T(v)$ for some $v \in V$ and thus equal to $\tau(v+im(T))$.

Theorem  (Universal mapping property for quotient spaces). Let $F$ be a field, $V,W$

vector spaces over $F$, $T : V → W$ a linear transformation, and $U \subset V$ a subspace.

If $U \subset ker(T)$, then there is a unique well-defined linear transformation

$\tau : V/U → W$ given by $\tau(v + U) = T(v)$.