Category of Banach spaces and bounded linear maps is not abelian

Question

Let $\mathbf{Ban}_{\infty}$ be the category of Banach spaces and bounded linear maps. I want to show that this category is not abelian. It is obviously additive, and every bounded linear map has kernels and cokernels in $\mathbf{Ban}_{\infty}$. So I must show that the first isomorphism theorem doesn't hold in $\mathbf{Ban}_{\infty}$ (I'm using Bass definition of abelian category). For every $T \in \mathscr{L}(E,F)$, the canonical morhpism is $$ \begin{array}{rcl} \widetilde{T} \colon F/ker(T) & \longrightarrow & \overline{im(T)}\\ \left[x\right] &\longmapsto & T(x) \end{array} $$ $\widetilde{T}$ is mono and epi in $\mathbf{Ban}_{\infty}$ (in fact, this holds in any quasi-abelian category). Any idea to show that $\widetilde{T}$ is not an iso of $\mathbf{Ban}_{\infty}$ in general? Maybe it's easier to take any bounded linear map epi and mono that is not an iso of $\mathbf{Ban}_{\infty}$: that can't happen in an abelian category. I don't care if you use another equivalent definition of abelian category (Peter Freyd's definition, for example), my goal is to show that $\mathbf{Ban}_{\infty}$ can't be abelian!

Answer 1

Well, the expression $\widetilde{T} \colon F/\ker(T) \to \overline{\operatorname{im}(T)}$ immediately suggests something: the image of $\widetilde{T}$ is obviously $\operatorname{im}(T)$, so $\widetilde{T}$ can't be surjective unless $\operatorname{im}(T)=\overline{\operatorname{im}(T)}$. So just take any example of a bounded linear map whose image is not closed, and $\widetilde{T}$ will not be an isomorphism.