On the AB5 condition

abelian-categories

I'm wondering on the present issue since few days: I'd like to understand why the so-called AB5 condition fails in general for a cocomplete (i.e. AB3) abelian category, more precisely I'd like you to highline the error I'm doing in the forthcoming thoughts.

Let $\cal A$ be a cocomplete abelian category, let $(u_i\colon A_i\hookrightarrow A)_{i\in I}$ be a direct family of subobjects of a fixed object $A\in{\cal A}$. Let $(\varinjlim A_i,(\mu_i)_{i\in I})$ be the direct limit of the family. Then we would have:
$$
\varinjlim_{i\in I}A_i=\sum_{i\in I}\mathop{\rm Im}\mu_i \qquad(1)
$$
and
$$
\alpha\circ\mu_i=u_i\quad\forall i\in I\qquad(2)
$$
where $\alpha\colon\varinjlim A_i\to A$ is the canonical morphisms.

Concerning (1), I took the canonical epimorphism $\pi\colon\bigoplus_{i\in I}A_i\to\varinjlim A_i$ defining the direct limit as a cokernel, then applied the right exact functor $\bigoplus$ on the family of the factorisations $A_i\to\mathop{\rm Im}\mu_i\to \varinjlim A_i$ of the $\mu_i$'s, to find
$$
\mathop{\rm Im}\pi=\varinjlim_{i\in I}A_i=\sum_{i\in I}\mathop{\rm Im}\mu_i
$$
as claimed. But (2) then tells me that each $\mu_i$ is a mono since $u_i$ is, whence $\mathop{\rm Im}\mu_i\cong A_i$ for all $i\in I$.

Now, I'm quite sure that I've already made a mistake so far, but in any case I'm asking you whether AB5 condition requires for the direct limit of a family of subobjects to be actually a subobject of $A$, not only an object isomorphic to a subobject of $A$. For, in the above confusing idea I should have proved that AB5 holds true in any cocomplete abelian category, which is not true in general.

Thank you!

Best Answer

Everything you wrote is fine - but the point is that in a general Abelian category, you cannot conclude from what you wrote, i.e. $A$ restricted to directed pieces being embedded, that the whole thing is embedded.

In other words: You just passed from the direct system $(A_i)_i$ to the isomorphic direct system $(\mathrm{Im}\mu_i)_i.$ Each of them comes with a system of monomorphisms to $A$, and the task/goal is to conclude that the induced map on colimit is a monomorphism.

But the task did not really get any easier by changing the situation up to isomorphism. The thing that creates an illusion that it does is that we tend to think of images as subsets of the target set, and so it seems that after taking union of these sets, one gets still a subset. This works well for module categories (where Ab5 holds) but fails in general abelian categories.

Related Solutions

Category Theory – Abelian Categories and Axiom (AB5)

I borrowed the ideas from the following books.

Abelian categories with application to rings and modules by Popescu, 1973.

Theory of categories by Mitchell, 1964.

Notations and Conventions We fix a Grothendieck universe $\mathcal{U}$. We consider only categories which belong to $\mathcal{U}$. Let $\mathcal{C}$ be a category. We denote by Ob($\mathcal{C}$) the set of objects of $\mathcal{C}$. Often, by abuse of notation, we use $\mathcal{C}$ instead of Ob($\mathcal{C}$). We denote by Mor($\mathcal{C}$) the set of morphisms of $\mathcal{C}$.

Let $f:X \rightarrow Y$ be a morphism of $\mathcal{C}$. We denote by dom($f$) the domain of $f$, i.e. $X$ = dom($f$). We denote by codom($f$) the codomain of $f$, i.e. $Y$ = codom($f$).

Definition Let $\mathcal{C}$ be a category. Let $X$ be an object of $\mathcal{C}$. Let $I$ be a small set. Let $(X_i)_I$ be a family of subobjects of $X$. If $(X_i)_I$ satisfies the following condition, $(X_i)_I$ is called a directed family of subobjects of $X$.

For any $i, j \in I$, there exists $k \in I$ such that $X_i \subset X_k$ and $X_j \subset X_k$.

Lemma 1 Let $\mathcal{A}$ be a cocomplete abelian category. Let $I$ be a small category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $A$ = colim $F$. For each $i \in I$, let $f_i:F(i) \rightarrow A$ be the canonical morphism. For each $i \in I$, let $A_i$ = Im($f_i$). Since $\mathcal{A}$ is cocomplete, $\sum A_i$ exists. Then $A = \sum A_i$.

Proof: Let $B = \sum A_i$. Let $m:B \rightarrow A$ be the canonical monomorphism. Since $A_i$ = Im($f_i$) for each $i \in I$, there exists $g_i:F(i) \rightarrow B$ such that $f_i = mg_i$. Let $u: i \rightarrow j$ be a morphism of I. Since $f_i = f_jF(u)$, $mg_i = mg_jF(u)$. Since $m$ is a monomorphism, $g_i = g_jF(u)$. Hence there exists $g:A \rightarrow B$ such that $g_i = gf_i$ for each $i$. Hence $mgf_i = mg_i = f_i$ for each $i$. Hence $mg = 1_A$. Hence $A \subset B$. Hence $A = B$. QED

Lemma 2 Let $\mathcal{C}$ be a cocomplete category. Let I be a small category. Let $\mathcal{C}^I$ be the category of functors: $I \rightarrow \mathcal{C}$. Then colim$: \mathcal{C}^I \rightarrow \mathcal{C}$ preserves colimits.

Proof: Let $\Delta: \mathcal{C} → \mathcal{C}^I$ be the diagonal functor, i.e. for each $X \in \mathcal{C}$ and for each $i \in I$, $\Delta(X)(i) = X$. Since colim is a left adjoint functor of $\Delta$, it preserves colimits(MacLane: Categories for the working mathematician, Chapter V, Section 5, Theorem 1, p.114). QED

Lemma 3 Let $\mathcal{A}$ be a cocomplete abelian category which satisfies (AB5). Let $X$ be an object of $\mathcal{A}$. Let I be a small filtered category. Let Sub($X$) be the category of subobjects of $X$. Let $F: I \rightarrow$ Sub($X$) be a functor.

Then $\sum F(i)$ = colim $F$.

Proof: For each $i \in I$, Let $u_i：F(i) \rightarrow$ colim $F$ be the canonical morphism. For each $i \in I$, Let $m_i：F(i) \rightarrow X$ be the canonical monomorphism. Since $(m_i)_I$ is a cocone, it induces a morphism $f$: colim $F \rightarrow X$. By (AB5), $f$ is mono. Hence we can regard colim $F$ as a subobject of $X$. Since $fu_i = m_i$ for each $i$, $F(i) \subset$ colim $F$.

Let $Z$ be a subobject of $X$. Let $r: Z \rightarrow X$ be the canonical monomorphism. Suppose $F(i) \subset Z$ for each $i$. Let $k_i: F(i) \rightarrow Z$ be the canonical monomorphism. Since $(k_i)_I$ is a cocone, it induces a morphism $g$: colim $F \rightarrow Z$. For each $i \in I$, $rgu_i = rk_i = m_i$. Hence $f = rg$. Hence colim $F \subset Z$. QED

Lemma 4 Let $\mathcal{A}$ be a cocomplete abelian category. Let $f:X \rightarrow Y$ be a morphism of $\mathcal{A}$. Let $I$ be a small set. Let $(X_i)_I$ be a family of subobjects of $X$. Then $\sum f(X_i) = f(\sum X_i)$.

Proof: For each $i \in I$, $X_i \subset \sum X_i$. Hence $f(X_i) \subset f(\sum X_i)$.

Let $Z$ be a subobject of $X$. Suppose $f(X_i) \subset Z$ for each $i$. Then $f^{-1}(f(X_i)) \subset f^{-1}(Z)$. Since $X_i \subset f^{-1}(f(X_i))$, $X_i \subset f^{-1}(Z)$. Hence $\sum X_i \subset f^{-1}(Z)$. Hence $f(\sum X_i) \subset f(f^{-1}(Z)) \subset Z$. QED

Lemma 4.5 Let $\mathcal{A}$ be a cocomplete abelian category. Let $I$ be a small category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $X$ = colim $F$. Let $(s_i: F(i) \rightarrow Y)_I$ be a cocone. Let $f:X \rightarrow Y$ be the morphism induced by the cocone. Then $f(X) = \sum s_i(F(i))$.

Proof: For each $i$, let $u_i: F(i) \rightarrow X$ be the canonical morphism. For each $i$, $fu_i = s_i$. Hence $f(u_i(F(i)) = s_i(F(i)) \subset f(X)$.

Let $Z$ be a subobject of $Y$. Suppose $s_i(F(i)) \subset Z$ for each $i$. For each $i$, $s_i$ induces $t_i: F(i) \rightarrow Z$. Since $(t_i: F(i) \rightarrow Z)_I$ is a cocone, it induces $g:X \rightarrow Z$. Let $m: Z \rightarrow Y$ be the canonical monomorphism. $mgu_i = mt_i = s_i$ for each i. Hence $f = mg$. Hence $f(X) \subset Z$. QED

Lemma 5 Let $\mathcal{A}$ be a cocomplete abelian category. Let $X$ be an object of $\mathcal{A}$. Let $I$ be a small set. Let $(X_i)_I$ be a family of subobjects of $X$. Then $\bigoplus X/X_i$ = $X/(\sum X_i)$.

Proof: For each $i \in I$, the following sequence is exact. $0 \rightarrow X_i \rightarrow X \rightarrow X/X_i \rightarrow 0$. By Lemma 2, colim preserves cokernels. Hence, colim $X_i \rightarrow X \rightarrow$ colim $X/X_i \rightarrow 0$ is exact. By Lemma 4.5, Im(colim $X_i \rightarrow X$) = $\sum X_i$. Hence colim $X/X_i$ = $X/(\sum X_i)$. QED

Lemma 5.4 Suppose the following is a pullback diagaram in an abelian category. $$\begin{matrix} A&\stackrel{f}{\rightarrow}&B\\ \downarrow&&\downarrow\\ C&\stackrel{h}{\rightarrow}&D \end{matrix} $$

Suppose the following sequence is exact. $0 \rightarrow C \stackrel{h}{\rightarrow}D \rightarrow E$

Then $0 \rightarrow A \stackrel{f}{\rightarrow}B \rightarrow E$ is exact.

Proof: Left to the readers.

Lemma 5.5 Consider the following commutative diagram with two horizontal exact sequences in an abelian category.

$X \rightarrow Y \rightarrow Z \rightarrow 0$

$0 \rightarrow X' \rightarrow Y' \rightarrow Z' \rightarrow 0$

Suppose the left square is a pullback. Then $Z \rightarrow Z'$ is mono.

Proof: We call s the above morphism $Z \rightarrow Z'$.

Let $r:T \rightarrow Z$ be a morphism such that sr = 0.

There exists the following pullback diagaram. $$\begin{matrix} P&\stackrel{u}{\rightarrow}&T\\ \downarrow&&\downarrow{r}\\ Y&\stackrel{}{\rightarrow}&Z \end{matrix} $$

By Lemma 5.4, $0 \rightarrow X \rightarrow Y \rightarrow Z’$ is exact.

Hence there exists $P \rightarrow X$ such that $P \rightarrow Y = P \rightarrow X \rightarrow Y$.

Hence $ru$ = 0. On the other hand, since a pullback of an epimorphism in an abelian category is epi(MacLane Proposition 2, p.199), $u$ is epi. Hence $r$ = 0. QED

Lemma 6 Let $\mathcal{A}$ be an abelian category. Let $f:X \rightarrow Y$ be a morphism of $\mathcal{A}$. Let $Z \subset Y$. Then $X/f^{-1}(Z)$ is canonically isomorphic to $f(X)/(f(X) \cap Z)$.

Proof:

Consider the following commutative diagram with two horizontal exact sequences.

Diagram

By Lemma 5.5, $X/f^{-1}(Z) \rightarrow f(X)/(f(X) \cap Z)$ is mono. Since $X \rightarrow f(X)$ is epi, $X \rightarrow X/f^{-1}(Z) \rightarrow f(X)/(f(X) ∩ Z)$ is epi. Hence $X/f^{-1}(Z) \rightarrow f(X)/(f(X) \cap Z)$ is epi. Hence $X/f^{-1}(Z) \rightarrow f(X)/(f(X) \cap Z)$ is an isomorphism. QED

Note If you are willing to accept Mitchell's embedding theorem, Lemma 6 will be trivial.

Lemma 7 Let $\mathcal{A}$ be a cocomplete abelian category. Suppose $\mathcal{A}$ has the following property.

Let $A$ be an object of $\mathcal{A}$. Let $(A_i)_I$ be a directed family of subobjects of $A$. Then, for every subobject $B$ of $A$, $(\sum A_i) \cap B = \sum (A_i \cap B)$.

Let $f:Y \rightarrow X$ be a morphism of $\mathcal{A}$. Let $(X_i)_I$ be a directed family of subobjects of $X$. Then, $f^{-1}(\sum X_i) = \sum f^{-1}(X_i)$.

Proof: By Lemma 6, for each $i$, $Y/f^{-1}(X_i)$ is canonically isomorphic to $f(Y)/(f(Y) \cap X_i)$.

Hence $\bigoplus Y/f^{-1}(X_i)$ is canonically isomorphic to $\bigoplus f(Y)/(f(Y) \cap X_i)$.

By Lemma 5, $Y/\sum f^{-1}(X_i)$ = $\bigoplus Y/f^{-1}(X_i)$.

Hence $Y/\sum f^{-1}(X_i)$ = $\bigoplus f(Y)/(f(Y) \cap X_i)$.

By Lemma 5, $\bigoplus f(Y)/(f(Y) \cap X_i)$ = $f(Y)/\sum (f(Y) \cap X_i)$.

By the assumption, $f(Y)/\sum (f(Y) \cap X_i)$ = $f(Y)/((\sum X_i) \cap f(Y))$.

By Lemma 6, $Y/f^{-1}(\sum X_i)$ is canonically isomorphic to $f(Y)/((\sum X_i) \cap f(Y))$.

Hence $Y/\sum f^{-1}(X_i)$ is canonically isomorphic to $Y/f^{-1}(\sum X_i)$.

Hence $f^{-1}(\sum X_i)$ = $\sum f^{-1}(X_i)$. QED

Lemma 7.3 Let $\mathcal{C}$ be a category. Let $X$ be an object of $\mathcal{C}$. Let Sub($X$) be the category of subobjects of $X$. Let $I$ be a small set. Let $(X_i)_I$ be a directed family of subobjects of $X$. Then there exists a preorder on $I$ making $I$ a filtered category and a functor $F: I \rightarrow$ Sub($X$) such that $F(i) = X_i$ for each $i \in I$.

Proof: Define $i \leq j$ if and only if $X_i \subset X_j$. QED

Lemma 7.5 Let $\mathcal{A}$ be an abelian category. Let $I$ be a small category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $i \in I$. Let $(i\downarrow I)$ be the coslice category under i. Let Sub($F(i)$) be the category of subobjects of $F(i)$. Then there exists a functor $G$: $(i\downarrow I) \rightarrow$ Sub($F(i)$) such that $G(u)$ = Ker($F(u)$) for each $u \in (i\downarrow I)$.

Proof:Clear.

Lemma 8 Let $\mathcal{A}$ be an abelian category. Let $I$ be a small filtered category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $i \in I$. Let $J$ = {$u \in$ Mor($I$); $i$ = dom($u$)}. Then (Ker($F(u))$)$_J$ is a directed family of subobjects of $F(i)$.

Proof: Let $(i\downarrow I)$ be the coslice category under i. $(i\downarrow I)$ is clearly a filtered category. Since $J$ = Ob($(i\downarrow I)$), the assertion follows immediately from Lemma 7.5. QED

Lemma 8.5 Let $\mathcal{A}$ be a cocomplete abelian category. Let $I$ be a small category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $S = \bigoplus_i F(i)$, where $i$ runs over every object of $I$. Let $m_i: F(i) \rightarrow S$ be the canonical monomorphism for each $i \in I$. Let $M$ = $\sum_u$ Im($m_i - m_jF(u)$), where $u$ runs over every morphism of $I$ and $i$ = dom($u$), $j$ = codom($u$). Let $\pi:S \rightarrow S/M$ be the canonical epimorphism. Let $f_i = \pi m_i$ for each $i \in I$. Then $S/M$ = colim $F$ with canonical morphisms $f_i: F(i) \rightarrow S/M$ for each $i \in I$.

Proof:Left to the readers.

Lemma 8.6 Let $I$ be a filtered category. Let $V$ be a non-empty finite subset of Ob($I$). Let $T$ be a finite subset of Mor($I$) such that dom($u$) $\in V$ and codom($u$) $\in V$ whenever $u \in T$. Then there exists $p \in$ Ob($I$) and a morphism $f_i: i \rightarrow p$ for each $i \in V$ with the following property.

For each $u:i \rightarrow j$ in $T$, $f_i = f_ju$.

Proof: There exists $q \in I$ such that there exists a morphism $g_i:i \rightarrow q$ for each $i \in V$. Let $u:i \rightarrow j$ in $T$. There exists $r_u \in I$ and a morphism $h_u:q \rightarrow r_u$ such that $h_ug_i = h_ug_ju$. There exists $r \in I$ such that there exists a morphism $r_u \rightarrow r$ for each $u \in T$.
Hence, for each $u:i \rightarrow j$ in $T$ there exist a morphism $g_{u, i}: i \rightarrow r$ and a morphism $h_{u, j}: j \rightarrow r$ such that $g_{u, i} = h_{u, j}u$.

For each $i \in V$, let $G_i$ be the set {$g_{u, i}: i$ = dom($u$), $u \in T$}, and let $H_i$ be the set {$h_{u, i}: i$ = codom($u$), $u \in T$}. Let $S_i = G_i \cup H_i$ for each $i \in V$. By the properties of a filtered category, we can assume that $S_i$ consists of one morphism $f_i$ with a common codomain $p$ for each $i \in V$. If $S_i$ is empty, we can assume that there exists a morphism $f_i:i \rightarrow p$ which has no condition.
QED

Lemma 9 Let $\mathcal{A}$ be a cocomplete abelian category. Suppose $\mathcal{A}$ has the following property.

Let $A$ be an object of $\mathcal{A}$. Let $(A_i)_I$ be a directed family of subobjects of $A$. Then, for every subobject $B$ of $A$, $(\sum A_i) \cap B = \sum (A_i \cap B)$.

Let I be a small filtered category. Let $F: I \rightarrow \mathcal{A}$ be a functor. For each $i$, let $f_i: F(i) \rightarrow$ colim($F$) be the canonical morphism. Then, for each $i$, Ker($f_i$) = $\sum$ Ker($F(u)$), where $u$ runs over every morphism such that $i$ = dom($u$).

Proof: We use the notations of Lemma 8.5. Let $T$ be a subset of Mor($I$). Let $M_T$ = $\sum_{u \in T}$ Im($m_i - m_jF(u)$), where $i$ = dom($u$), $j$ = codom($u$). Then $M = \sum_T M_T$, where T runs through all finite subsets of Mor($I$). Hence, by Lemma 8.5 and Lemma 7, Ker($f_i$) = $m_i^{-1}(M)$ = $\sum_T m_i^{-1}(M_T)$, where T runs through all finite subsets of Mor($I$). It suffices to prove: For each finite subset $T$ of Mor($I$), $m_i^{-1}(M_T) \subset$ Ker($F(u)$) for some $u \in$ Mor($I$) such that $i$ = dom($u$).

Let $V$ be the set of $k \in I$ such that $k$ = $i$ or $k$ = dom($u$) or $k$ = codom($u$) for some $u \in T$.

Since $I$ is filtered, by Lemma 8.6, there exists $p \in$ Ob($I$) and a morphism $v_k: k \rightarrow p$ for each $k \in V$ with the following property.

For each $u \in$ Mor($I$) such that k = dom($u$) $\in V$ and $j$ = codom($u$) $\in V$, $v_k = v_ju$.

We define $f:S \rightarrow F(p)$ as follows. Let $k$ be any object of $I$. If $k \in V$, $fm_k = F(v_k)$, otherwise $fm_k = 0$.

For each $u \in T$, let $k$ = dom($u$), $j$ = codom($u$). Then $f(m_k - m_jF(u))$ = $F(v_k) - F(v_j)F(u)$ = $0$. Hence, by Lemma 4, $f(M_T)$ = $0$. Since $m_i(m_i^{-1}(M_T)) \subset M_T$, $0$ = $f(m_i(m_i^{-1}(M_T)))$ = $F(v_i)(m_i^{-1}(M_T))$. Hence $m_i^{-1}(M_T) \subset$ Ker($F(v_i)$) as required. QED

Proposition 1 Let $\mathcal{A}$ be a cocomplete abelian category. Suppose $\mathcal{A}$ has the following property.

Let $A$ be an object of $\mathcal{A}$. Let $(A_i)_I$ be a directed family of subobjects of $A$. Then, for every subobject $B$ of $A$, $(\sum A_i) \cap B = \sum (A_i \cap B)$.

Then $\mathcal{A}$ satisfies (AB5).

Proof: Let $I$ be a small filtered category. By Lemma 2, colim$: \mathcal{A}^I \rightarrow \mathcal{A}$ preserves colimits. In particular, it preserves cokernels. Hence it is right exact. It suffices to prove that it preserves monomorphisms.

Let $f: F \rightarrow G$ be a monomorphism of $\mathcal{A}^I$. Let $K$ = Ker(colim($f$)). For each i, let $u_i: F(i) \rightarrow$ colim $F$ be the canonical morphism. Let $A_i$ = $u_i(F(i))$ for each i. Since $I$ is a filtered category, $(A_i)_I$ is a directed family of subobjects of colim $F$. By Lemma 1, colim $F$ = $\sum A_i$. By the assumption, $K$ = $(\sum A_i) \cap K = \sum (A_i \cap K)$. Suppose $K \neq 0$. There exists $k \in I$ such that $A_k \cap K \neq 0$. Since $A_k$ = Im($u_k$), $u_k^{-1}(A_k \cap K) \neq 0$. Let $M = u_k^{-1}(A_k \cap K)$. Then $u_k(M) \neq 0$. For each i, let $v_i: G(i) \rightarrow$ colim $G$ be the canonical morphism. $v_k(f_k(M))$ = (colim $f$)($u_k(M)$) = (colim $f$)($A_k \cap K$) $\subset$ (colim f)($K$) = $0$. Hence $f_k(M) \subset$ Ker($v_k$). By Lemma 9, $f_k(M) \subset \sum$ Ker($G(t)$), where $t$ runs over every morphism such that $k$ = dom($t$).

By Lemma 8 and the assumption, $f_k(M)$ = $\sum$ (Ker($G(t)$) $\cap f_k(M)$). Since $f_k$ is mono, $M$ = $f_k^{-1}(f_k(M))$. By Lemma 7, $M$ = $f_k^{-1}(f_k(M))$ = $f_k^{-1}(\sum$ (Ker($G(t)$) $\cap f_k(M)))$ = $\sum f_k^{-1}$(Ker($G(t)$ $\cap f_k(M))$. For each morphism $t: k \rightarrow j$, Let $N_t = f_k^{-1}$(Ker($G(t)$) $\cap f_k(M))$. Then $G(t)f_k(N_t) = 0$. Since $G(t)f_k = f_jF(t)$, $f_jF(t)(N_t) = G(t)f_k(N_t) = 0$. Since $f_j$ is mono, $F(t)(N_t) = 0$. Hence, by Lemma 9, $u_k(N_t) = 0$. Hence, by Lemma 4, $u_k(M) = u_k(\sum N_t) = \sum u_k(N_t) = 0$. This is a contradiction. QED

Proposition 2 Let $\mathcal{A}$ be a cocomplete abelian category satisfying (AB5). Let $A$ be an object of $\mathcal{A}$. Let $(A_i)_I$ be a directed family of subobjects of $A$. Then, for every subobject $B$ of $A$, $(\sum A_i) \cap B = \sum (A_i \cap B)$.

Proof: Let $C = \sum A_i$. For each i, we have the following exact sequence.

$0 \rightarrow A_i \cap B \rightarrow A_i \rightarrow C/(C \cap B)$

By (AB5), Lemma 7.3 and Lemma 3, we get the following exact sequence.

$0 \rightarrow \sum (A_i \cap B) \rightarrow C \rightarrow C/(C \cap B)$

Hence $(\sum A_i) \cap B = \sum (A_i \cap B)$. QED

Regarding condition $AB5$

I am not sure whether I understand your question, but for abelian categories, AB5 is equivalent to the fact that filtered colimits preserve finite limits.

First of all, a filtered colimit functor, or any colimit functor, on an abelian category is always right exact (since it is a left adjoint to the diagonal embedding/constant diagram functor). It follows that when a filtered colimit functor is not exact, i.e. AB5 fails, the failure of exactness is due to the failure of left exactness, i.e. failure to preserve kernels.

(In a sense, this occurs in nature quite often, i.e. take $\mathcal{A}=(\mathrm{Mod-}R)^{op}$. Then this is typically not an AB5-category since inverse limits are typically not exact in $\mathrm{Mod-}R$.)

On the other hand, in an AB5-abelian category, filtered colimit functors preserve all finite limits. The reason is that one can represent all limits as a composition of simpler ones. Typically, all limits are obtained just as equalizer of pair of morphisms between certain products (whose form depends only on the diagram shape), and in the case of finite limits in an Abelian category, any finite limit is obtained as a kernel of a map between some finite biproducts, i.e. coproducts. But again, filtered colimit functor always preserves coproducts (it is a left adjoint), so preserving finite limits again comes down to preserving kernels, which is guaranteed by exactness.

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Related Solutions

Category Theory – Abelian Categories and Axiom (AB5)

Regarding condition $AB5$

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