If $\text{ker }f$ lies in $\mathcal{C}$ then $\text{ker }f$ is the same in $\mathcal{A}$.

I have seen this used as an argument in my textbook:

Set-up:

Assume $\mathcal{C}$ is a full subcategory of an Abelian category $\mathcal{A}$. Let $f: A \rightarrow B$ be a morphism in $\mathcal{C}$.

If $\text{ker }f$ lies in $\mathcal{C}$ then $\text{ker }f$ is the same in $\mathcal{A}$.

(They also assumed in the textbook that the zero object in $\mathcal{A}$ lies in $\mathcal{C}$ and the direct sum of $A,B \in \text{Obj} \mathcal{C}$ lies in $\mathcal{C}$ and the $\text{coker} f$ lies in $\mathcal{C}$.)

Is this true? And how do I prove it?

I think my problem is my understanding of the definition of the kernel. In the textbook the kernel of $f$ in $\mathcal{C}$ should have the property: Let $i:K\rightarrow A$ be the kernel of $f$ then

For every $g: X \rightarrow A$ with $fg = 0$ we have a unique $\theta: X \rightarrow K$ with $i\theta = g$.

What is $g$? I have assumed $g$ is a morphism in the category $\mathcal{C}$ and thus I cannot use $i$ as the kernel of $f$ in $\mathcal{A}$ since I only have the properties for some morphisms $g$.

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In general, if $i : C \subseteq A$ is a full subcategory and $X$ is a diagram in $C$ such that $iX$ has a limit in $A$ which is of the form $iL$ for some $L \in C$, then $X$ has limit $L$. The proof is trivial, nothing happens at all. Morally, if the universal property of the limit diagram is witnessed by all objects of $A$, then a fortiori by all objects of $C$.

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Never heard about limits and limit diagram. Is it possible to explain it without using limit and diagrams? – postguest12 Jan 2 '14 at 14:00
Then replace $X$ by a morphism and limit by kernel. – Martin Brandenburg Jan 2 '14 at 14:01

Hint: The way to go is to assume that there is a kernel in $\mathcal A$ and a kernel in $\mathcal C$, say $(\ker f)_{\mathcal A}$ and $(\ker f)_{\mathcal C}$. Now try to prove that these are isomorphic.

To do that you will use the universal property of $(\ker f)_{\mathcal A}$ on the natural map $(\ker f)_{\mathcal C} \to A$.

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This is not "the way to go", but rather "a way to go" and morevover one of the longest ones. Also notice that it is formally not correct because - a priori - $(\ker(f))_C$ has no reason to exist. – Martin Brandenburg Jan 2 '14 at 0:20

I think I found a counterexample to the dual problem:

Let $\mathcal A$ be the category of Abelian groups and $\mathcal C$ be the full subcategory of free groups of finite rank (i.e. $\Bbb Z^n$'s). It is closed under direct sum and taking kernels, but is not closed under cokernels (as e.g. the cokernel of $\Bbb Z\overset{x\mapsto 2x}\longrightarrow\Bbb Z$ is not in $\mathcal C$).

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Ah, didn't consider the part 'if $\ker f$ lies in $\mathcal C$'. Then of course, the other two answers apply. – Berci Jan 2 '14 at 0:28