Canonical morphism is always monic and epic? I'm reading Manin & Gelfand's book Methods of Homological Algebra and I came to something I can't prove after several tries.
It's on page 113, II.5.11.(b) Commentary on the Axiom A4 (of Abelian Categories). There it's stated that in every category with axioms A1-A3 and such that kernels and cokernels exists, then the canonical morphism is always a monomorphism and an epimorphism.
I've been working on this but can't obtain anything. What I found is the following:
With the standard procedure, starting with a morphism $\varphi: a \to b \in \mathcal{C},$ we can obtain maps $h : I \to b$ such that $\varphi = h \circ {\tt coker}({\tt ker}\ \varphi),$ $h': a \to J$ such that $\varphi = {\tt ker}({\tt coker}\ \varphi) \circ h'$ and $\ell :I \to J$ such that $h = {\tt coker}({\tt ker}\ \varphi)\circ \ell.$ Here, $I$ is the target of ${\tt coker}({\tt ker}\ \varphi)$ and $J$ is the source of ${\tt ker}({\tt coker}\ \varphi).$
Clearly, $\ell$ is the canonical morphism associated to $\varphi.$ Further properties of the morphisms above are:

*

*${\tt ker}\ h = {\tt ker}\ \ell,$

*${\tt coker}\ h' = {\tt coker}\ \ell,$

*${\tt ker}\ \varphi = {\tt ker}\ h',$

*${\tt coker}\ \varphi = {\tt coker}\ h,$

*${\tt ker}({\tt coker}({\tt ker}\ \varphi)) = {\tt ker}\ \varphi,$

*${\tt coker}({\tt ker}({\tt coker}\ \varphi)) = {\tt coker}\ \varphi.$
Using all this properties I can't prove that ${\tt ker}\ \ell = 0,{\tt coker}\ \ell = 0.$ Could someone help me with this?
Thanks in advance!
 A: If I understand the question properly, the answer is: It is not true that the induced morphism between the coimage and image is always a mono- and an epimorphism.
An example is the category $C$ of complete locally convex Hausdorff spaces and continuous linear maps. Kernels in $C$ are the usual null sets with the subspace topology and monomorphisms are just the injective continuous linear maps.The cokernel of $f:X\to Y$ is more subtle, it is the completion of $Y/\overline{f(X)}$, where you first need the closure of the range to obtain a Hausdorff space $Y/\overline{f(X)}$ (in the quotient topology of $Y$) which however need not be complete (it is complete if $Y$ is metrizable, but in general quotients of complete spaces needn't be complete, examples can be seen here https://mathoverflow.net/questions/57654) so that you have to take the completion.
The coimage of $f$ is thus the completion of $X/$ker$(f)$ and the image is the closure $\overline{f(X)}$ of the range.
If now $f:X\to Y$ is a morphism with closed range $f(X)$ such that $Z=X/$ker$(f)$ is not complete then the induced morphism is the unique continuous extension to the completion of the canonical map $Z\to f(X)$ which fails to be injective.
All this is discussed in an article of Kopylov and Wegner On the notion of a semi-abelian category in the sense of Palamodov (here is a link to an easily accissible version https://arxiv.org/abs/1406.6804).
