# Two versions for 1st Isomorphism Theorem for Banach Spaces

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?

• I'd rather say that open mapping theorem implies theorem 2. – Eclipse Sun Jan 28 at 17:19
• @EclipseSun It does (I have just demonstrated that here), I just wanted to make sure that Theorem 1 is connected to Theorem 2, since they are very similar, But turns out one is stronger. – pitariver Jan 28 at 17:26
• What if $x\ne0$ but $Tx=0$? Then the second inequality in your proof is plainly wrong... – Vladimir Jan 28 at 17:26
• @Vladimir I fixed the norm equivalence to only on the quotient. Thanks for pointing that out. – pitariver Jan 28 at 17:59