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Denote by $\mathbf{Ban}$ the category of Banach spaces and bounded linear maps and by $\mathbf{Banc}$ the subcategory of Banach spaces and linear contractions. The isomorphisms of $\mathbf{Ban}$ are the bounded linear bijections (open mapping theorem) and in $\mathbf{Banc}$ they are the isometric (linear) isomorphisms (easy computation). $\mathbf{Ban}$ is additive and has all finite limits and finite colimits, so we can speak of (short) exact sequences.

proposition: the short five lemma holds in $\mathbf{Ban}$. proof: courtesy of the open mapping theorem, the diagram-chase proof for abelian groups transfers verbatim.

Now for my question:

Q: does the five short lemma also hold in $\mathbf{Banc}$?

Since $\mathbf{Banc}$ is not additive, a few words about the definitions. The crucial thing to notice is that kernels and cokernels in $\mathbf{Ban}$ are also kernels and cokernels in $\mathbf{Banc}$. For the sake of illustration, take the case of cokernels. In $\mathbf{Ban}$ they are given by the quotient map $\pi:B\to B/M$ with $M$ a closed subspace of $B$. If $T$ is a bounded linear map whose kernel contains $M$, then it factors uniquely through $\pi$ and the norm of the factorization equals the norm of $T$. But this is precisely what is needed for $\pi$ to be a cokernel in $\mathbf{Banc}$.

So now, we have two short exact sequences connected by maps $g$, $f$ and $h$ with $g$ and $h$ isometric isomorphisms and $f$ a linear contraction. The short five lemma holds if $f$ is an isometric isomorphism -- is there a way to draw diagrams? In any case, just look at short five lemma.

Start by noticing that by the above proposition $f$ must be a bijection and thus it has an inverse. What we need to prove is that this inverse is contractive. Alternatively, since we already know that $f$ is injective, the following easy result offers another route.

proposition: a map $f$ is a quotient iff it is surjective on open unit balls iff it is dense on unit balls.

Applying the above and chasing down an element of the open unit ball of the codomain of $f:B\to B'$, we get is that there is a $b$ in the open unit ball of $B$ and an $m$ in the kernel of the quotient in the top short exact sequence such that $f(b + m) = b'$. But of course, this does not get us very far.

At this point, I am beginning to suspect that the lemma does not hold, but I have been unable to rig a counterexample. The hypothesis is very strong and rules out the "usual" gang of suspects.

regards, G. Rodrigues

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up vote 10 down vote accepted

How about considering $\mathbb R^2$ with $l^1$ and $l^2$ norms. The restriction to the $x$-axis is an isometry. The quotient onto the $y$-axis, also an isometry. But not an isometry on the whole space.

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Simple, short and sweet. Thanks. – G. Rodrigues May 9 '11 at 1:18

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