# Category of Banach spaces and bounded linear maps is not abelian

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!

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.