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Let $f: A \to B$ and $g: A \to C$ be two monomorphisms in an abelian category. By definition, $B/A:= \operatorname{coker}(f)$, $C/A:= \operatorname{coker}(g)$.

If $h:B\to C$ is another morphism such that $h\circ f=g$, then by definition of cokernel, there exists a unique map $h':B/A \to C/A$ making the diagram commute.

My question is, does $h$ monic/epi imply $h'$ monic/epi? How to prove it without invoking Mitchell's embedding theorem(which requires the category to be small)?

In the category of $R$-modules, this is true(so by the embedding theorem this should be true for small abelian categories). But I want to know how to use an element-free argument to prove it(for categories that are big or small).

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  • $\begingroup$ Should not this work by snake lemma in abelian category? Though you may eventually quote embedding theorem, snake lemma is in general proved in abelian group category and then quote the embedding and another theorem along the way. $\endgroup$ – user45765 Jul 11 '17 at 15:45
  • $\begingroup$ @user45765 I know a proof of snake lemma without invoking the embedding theorem in MacLane's book(Around page 204, by his diagram chasing lemmas. The snake lemma also seems to be true for big abelian categories). Could you show me how to use the snake lemma to prove this? $\endgroup$ – No One Jul 11 '17 at 15:47
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    $\begingroup$ I recalled my version Diagram chasing lemma in Rotman's homological algebra requires 2 theorems one of which is embedding theorem. Anyway, you know it is abelian category. You can write Ker as 0 map. Then it will look like exact sequence $0\to A\to B\to Coker_1\to 0, 0\to A\to C\to Coker\to 0$ in abelian group category. There is morphism $B\to C$, $1_A:A\to A$. This induces map on $Coker_1\to Coker_2$ by exactness and commutivity of first square. Then apply snake lemma. Then you see $B\to C$ monic/epic implies coker level's monic/epic. $\endgroup$ – user45765 Jul 11 '17 at 15:52
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You don't really need the snake lemma for this.

Your situation can be described by the following diagram $$\require{AMScd} \begin{CD}0 @>>> A @>{f}>> B @>{\operatorname{coker}(f)}>> \frac{B}{A} @>>> 0 \\ & @V_{id_A}VV & @VV{h}V *& @VV{h'}V\\ 0 @>>> A @>>{g}>C @>>{\operatorname{coker}(g)}> \frac{C}{A} @>>> 0 \end{CD}$$ where both rows are short exact sequences. Then because the kernels are isomorphic, the square $*$ is a pullback (see for example this question), and pullbacks in abelian categories reflect monomorphisms, so if $h$ is a mono, then so is $h'$.

The case where $h$ is an epimorphism is easier : since cokernels are epimorphisms as well, $h'\circ \operatorname{coker}(f)=\operatorname{coker}(g)\circ h$ is an epimorphism, and thus $h'$ is an epimorphism.

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I will illustrate a method of generalizing diagram chasing arguments to arbitrary abelian categories. So, let us start with the following argument in the case of abelian groups: suppose $h$ is monic, and we want to show $h'$ is monic. It suffices to show $\ker(h') = 0$. Thus, suppose we have $\bar b \in B / A$ such that $h'(\bar b) = 0$. Then $\bar b$ has some preimage $b \in B$. Now, we know that $h(b) \in \mathrm{im}(g)$ by the assumption that $h'(\bar b) = 0$. Thus, choose $a \in A$ such that $g(a) = h(b)$ (which happens to be unique). Then $h(f(a)) = g(a) = h(b)$, so $f(a) = b$ since $h$ is monic. This implies that $\bar b = \overline{f(a)} = 0$.

To generalize this to abelian categories, suppose you have a test object $U$ and an element $\bar b \in (B/A)(U) := \mathrm{Hom}(U, B/A)$ such that $h'(\bar b) := h' \circ \bar b = 0$ in $(C/A)(U)$. Then there is some epimorphism $\phi : V \to U$ and a "section" $b \in B(V)$ such that $\phi^* \bar b := \bar b \circ \phi \in (B/A)(V)$ is equal to the projection of $b$. (In particular, $V := U \times_{B/A} B$ works, with $\phi$ and $b$ being the projections to $U$ and $B$ respectively.) By replacing $U$ with $V$ and $\bar b$ with $\phi^* \bar b$, we may assume $V = U$ and $\bar b$ is equal to the projection of $b$ (since the pullback by an epimorphism is injective). Now, since the projection of $h(b)$ to $(C/A)(U)$ is zero, this implies that $h(b)$ factors uniquely through $g$; let $a \in A(U)$ be the factorization, so that $g(a) = h(b)$. Then $h(f(a)) = g(a) = h(b)$, so since $h$ is monic, this implies $f(a) = b$. But this implies $\bar b = 0$. Therefore, we have shown that the zero map $0 \to B$ satisfies the universal property of $\ker(h')$, so $\ker(h') = 0$.

(So, the idea is: whenever the diagram-chasing argument makes use of the surjectivity of some map, translate that into replacing the test object $U$ by some other test object $U'$ with an epimorphism $U' \to U$ such that the section in question lifts, and pulling back all "sections" $X(U)$ to $X(U')$ which doesn't change equality of sections. And of course, if the diagram-chasing argument makes use of the injectivity of some map, that translates into directly using the fact that the corresponding map is monic. I find it helpful to think of the case where you are working on a category of sheaves of abelian groups, and then repeatedly refining a cover of the original open set $U$ over which sections are defined whenever we need to use the surjectivity of a sheaf morphism.)

(And if the problem were to show some map $\alpha : X \to Y$ is epic, then the translation of the diagram-chasing argument would show: for each test object $U$ and $y \in Y(U)$, there is an epimorphism $V \to U$ and "preimage" $x \in X(V)$ such that $\alpha(x)$ is the pullback of $y$. Then, applying this in the special case $y = \mathrm{id}_Y \in Y(Y)$, we would get that $\alpha \circ x$ is equal to the epimorphism $V \to Y$, which implies $\alpha$ is an epimorphism.)

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