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I am looking for verification of the following:

Let $X$ be a Banach space. We will show that if $M, N$ and $M+N$ are closed subspaces of $X$, then $$\dfrac{M+N}{N} \cong \dfrac{M}{M \cap N},$$ using the map $\phi : M \to \dfrac{M+N}{N}$, where $x \mapsto x + N$. For any $x,y \in M$, we have that, $$\phi(x+y) = (x+y) + N = (x + N) + (y + N) = \phi(x) + \phi(y)$$ Also, for any $\lambda \in \mathbb{F}$, we have, $$\phi(\lambda x) = (\lambda x) + N = \lambda(x + N) = \lambda\phi(x)$$ Then: \begin{align*} \ker(\phi) &= \left\{ m \in M : \phi(m) = e_{\frac{M+N}{N}} \right\} \hspace{1cm}\text{should this be sent to 0 instead of the "identity"?}\\ &= \left\{ m \in M : m + N = N \right\} \\ &= \left\{ m \in M : m \in N \right\} \\ &= M \cap N \end{align*} Then, we see that $\phi$ is a surjection because $$(m+n) + N = m + N \in \dfrac{M+N}{N}$$ which is $\phi(m)$. Then, we use the first isomorphism theorem, that is, for any operator $\phi: X \to Y$ between Banach spaces, $$X / \ker(\phi) \cong \text{im} \phi \iff \text{im}\phi \space\ \text{is closed in} \space\ Y$$ So that here, $X = M$, $\ker(\phi) = M \cap N$, and $Y = \text{im}(\phi) = \dfrac{M+N}{N}$, so that the result follows.

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    $\begingroup$ Please make explicit what it is that you're actually asking here. $\endgroup$ – Omnomnomnom Mar 25 '17 at 14:43
  • $\begingroup$ Proof verification, was unsure if I proved it correctly. $\endgroup$ – Dragonite Mar 25 '17 at 14:45
  • $\begingroup$ More so cautious to whether or not I missed something in the proof, did I show everything that needs to be shown here? Or does this proof basically mirror the one from intro abstract algebra courses? $\endgroup$ – Dragonite Mar 25 '17 at 14:49
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    $\begingroup$ To answer your question in the middle: indeed, $e_{(M+N)/N} = 0$. $0$ is the identity with respect to the group operation of addition. $\endgroup$ – Omnomnomnom Mar 25 '17 at 14:52
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Indeed, your proof should mirror that of the second isomorphism theorem. In fact, this statement is (almost) an instance of the second isomorphism theorem for modules, except for the fact that our $\phi$ (and eventually our isomoprhism) must also be continuous.

Your proof is correct except for two things:

  • You should write $0$ instead of $e$ (although your version is also technically correct)
  • You should note that $\phi$ is continuous, as is the induced map on $X/ \ker \phi$.
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  • $\begingroup$ Thank you, that makes sense. $\endgroup$ – Dragonite Mar 25 '17 at 15:06

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