I have encountered two different versions for the First Isomorphism Theorem in the context of Normed/Banach spaces, I wanted to ask whether one implies the other.
Theorem 1: Let $X,\,Y$ be normed spaces and suppose $T:X \rightarrow Y$ is a linear map such that $T(B_X)=B_Y$ where $B_X,B_Y$ are the corresponding open unit balls. Then $\frac{X}{ker(T)} \simeq Y$ are isometrically isomorphic.
Let me note that this version can be proved directly from definitions without using some sort of heavy hammer. Also note that $T$ is automatically bounded and surjective and of norm $1$.
Theorem 2: Let $X, Y$ be Banach spaces. $T:X \twoheadrightarrow Y$ A bounded surjective operator. Then $\frac{X}{ker(T)} \simeq Y$ are isomorphic (i.e. have a linear homeomorphism between them).
I claim that Theorem 1 implies Theorem 2: Let $X, Y, T$ be as above. Define a new norm $\Vert \cdot \Vert_T$ on $X$ as follows:
$$ \Vert x \Vert_T = \Vert Tx \Vert_Y \leq \Vert T \Vert \Vert x \Vert_X$$
So this inequality must also hold in the quotient space(I use the same notation for the quotient norm):
$$\forall x \; \Vert x + ker(T) \Vert_T \leq \Vert T \Vert \Vert x + ker(T) \Vert_X$$
But by the Open Mapping Theorem: $T$ is an open map and thus There exists $r > 0$ s.t. $\Vert Tx \Vert_Y \leq r \implies \Vert x + ker(T) \Vert_X \leq 1$. So for $x \notin ker(T)$, we have $$ \Vert T\big( r \frac{x}{\Vert Tx \Vert_Y}\big) \Vert_Y \leq r \implies \Vert r \frac{x}{\Vert Tx \Vert_Y} + ker(T) \Vert_X \leq 1$$
$$\implies \Vert x + ker(T) \Vert_X \leq \frac{1}{r}\Vert x \Vert_T$$
So this gives us in the quotient space:
$$\forall x \; \Vert x + ker(T) \Vert_X \leq \frac{1}{r} \Vert x + ker(T) \Vert_T$$ From these two inequalities we conclude that $\Vert \cdot \Vert_X,\,\Vert \cdot \Vert_T$ are equivalent norms on $\frac{X}{ker(T)}$ (the identity is a homeomorphism). Note that now $T(B_{\Vert \cdot \Vert_T}) = B_Y$ as a direct consequence of the definition of $\Vert \cdot \Vert_T$. So by Theorem 1, $(\frac{X}{kerT}, \Vert \cdot \Vert_T) \simeq (Y, \Vert \cdot \Vert_Y)$ are isometrically isomorphic and thus isomorphic, but the norms are equivalent and we are done.
is the preceding proof correct?