Inverse Mapping Theorem implies Open Mapping Theorem

I need to assume that the Inverse Mapping Theorem is true and deduce from it the Open Mapping Theorem. It is immediate to show that the OMP implies IMP.

The converse however seems harder. This is since somehow we should use IMP on a bounded linear function $T$ which we should assume is surjective, but not injective. (If it is injective then we can easily see that OMP follows.)

For concreteness let $T: X \to Y$ be a surjective continuous linear map between Banach spaces. To actually use the statement, I suppose we should restrict $X$ to some set $M$ such that $T$ there is a bijection. Hence by the IMP, $T$ is indeed open. Other than that I don't see how can we show it is still open in the complement of $M$.

Can someone help me prove this implication?

Solution 1:

We are going to need the following elementary lemma:

Lemma: Let $X$ be a normed space and $M$ a closed subspace. Then the quotient map $q:X\to X/M$ is an open map.

Proof of lemma without using the open mapping theorem: Let $U\subset X$ be an open set. To show that $q(U)$ is open, let $x\in U$; by $U$'s openness find $r>0$ such that $B(x,r)\subset U$. Now let $x'+M\in X/M$ be such that $\|(x+M)-(x'+M)\|<r$. Then $\|(x-x')+M\|<r$, i.e. $\inf_{m\in M}\|x-x'+m\|<r$. Since this infimum is strictly smaller than $r$, we obtain some $m\in M$ such that $\|x-x'+m\|<r$. But this shows that $x'-m\in B(x,r)\subset U$, so $x'-m\in U$ and thus $q(x'-m)\in q(U)$ and note that $q(x'-m)=q(x')=x'+M$. We have just shown that $B(x+M,r)\subset q(U)$, proving that $q(U)$ is open.

Assume now that we know the inverse mapping theorem, i.e. if $T:X\to Y$ is a bounded linear operator between Banach spaces that is a bijection, then it's inverse $T^{-1}:Y\to X$ (which is obviously linear) is also bounded.

Let $T:X\to Y$ be a bounded surjective map between Banach spaces. We shall show that $T$ has to be an open map. To do that, let $M=\ker(T)$. This is a closed subspace of $X$ so we can consider the quotient space $X/M$. Note that $T$ induces a bounded linear operator $\bar{T}:X/M\to Y$ by $\bar{T}(x+M)=Tx$. Let's check this: Indeed $\bar{T}$ is well-defined: if $x+M=x'+M$, then $x-x'\in M=\ker(T)$ so $Tx=Tx'$. Linearity is obvious. For boundedness, note that for any $m\in M$ we have $$\|\bar{T}(x+M)\|=\|Tx\|=\|T(x+m)\|\le\|T\|\cdot\|x+m\|$$ and since this is true for any $m\in M$ we can take the infimum over $m$ in the above inequality to conclude that $$\|\bar{T}(x+M)\|\le\|T\|\cdot\inf_{m\in M}\|x+m\|=\|T\|\cdot\|x+M\|_{X/M}$$ It is also obvious that $T$ is bijective: if $T(x+M)=0$ then $Tx=0$ so $x\in M$, i.e. $x+M=M=0_{X/M}$, showing injectivity. If $y\in Y$ then find $x\in X$ with $Tx=y$, by $T$'s assumed surjectivity; thus $y=\bar{T}(x+M)$, proving surjectivity of $\bar{T}$. Apply the inverse mapping theorem (note that $X/M$ is a Banach space itself, since $M$ is a closed subspace of a Banach space!) to $\bar{T}$ to obtain a bounded linear operator $S:Y\to X/M$ with the property that $S\bar{T}=\text{id}_{X/M}$ and $\bar{T}S=\text{id}_Y$.

Now let $U\subset X$ be an open set. Then $U+M$ is open in $X/M$, by our lemma. Now since $S$ is continuous, we have that $S^{-1}(U+M)$ is open in $Y$, by definition of continuity. But since $S^{-1}=\bar{T}$, we have that $S^{-1}(U+M)=\bar{T}(U+M)=T(U)$: let's prove the last equality; if $x\in U$, then $Tx=\bar{T}(x+M)\in\bar{T}(U+M)$, showing that $T(U)\subset\bar{T}(U+M)$. For the other inclusion, if $x\in X$ with $x+M\in U+M$, then find $x'\in U$ such that $x+M=x'+M$. But then $\bar{T}(x+M)=\bar{T}(x'+M)=Tx'\in T(U)$.

Since $\text{open}=S^{-1}(U+M)=T(U)$, we are done.