Category of abelian groups is not equivalent to its opposite I want to show that $\operatorname{Ab}$ is not equivalent to $\operatorname{Ab}^{op}$.
Is it sufficient to say that $\operatorname{Hom}(\mathbb{Z}, \mathbb{Z}_2)$ is not isomorphic to $\operatorname{Hom}(\mathbb{Z}_2, \mathbb{Z})$ ? If yes then I don't understand why.
 A: Here is a simpler argument: in any abelian category with both infinite coproducts and infinite products, there is a natural map
$$\bigoplus_{i \in I} X_i \to \prod_{i \in I} X_i$$
from the infinite coproduct to the infinite product. In $\text{Ab}$ this map is a monomorphism but almost never an epimorphism (this happens iff all but finitely many of the $X_i$ are nonzero), while in $\text{Ab}^{op}$ this map is, dually, an epimorphism but almost never a monomorphism.
To get some intuition for why coproducts in $\text{Ab}^{op}$ are "larger" than products, by Pontryagin duality $\text{Ab}^{op}$ is equivalent to the category of compact Hausdorff abelian groups. In this category products are cartesian products but coproducts are much more complicated: one takes the Bohr compactification of the algebraic coproduct equipped with a suitable topology. This compactification adds a lot of stuff; for example the infinite direct sum $\bigoplus_{\mathbb{N}} S^1$ of countably many copies of $S^1$ is the Pontryagin dual of $\mathbb{Z}^{\mathbb{N}}$, which is some very weird complicated thing.
Also a much more general result is true: $\text{Ab}$ is a locally presentable category, and it's a theorem due to Gabriel and Ulmer that if the opposite of a locally presentable category is locally presentable, then it is a preorder. So the opposite $C^{op}$ of a nontrivial (not a preorder) locally presentable category $C$ is never locally presentable, and in particular is never equivalent to $C$.
A: No, it's not enough (as @Aphelli and @milore have pointed out). Consider the category $C$ with two objects $X_0$, $X_1$ and a single arrow $a: X_0 \to X_1$. Certainly $\operatorname{Hom}(X_0, X_1)$ and $\operatorname{Hom}(X_1, X_0)$ are not isomorphic, but $C$ and $C^{\operatorname{op}}$ are equivalent (and in fact isomorphic) by exchanging $X_0$ and $X_1$.
A: The category of abelian groups has the following property: If
$$(0 \to A_i \to B_i \to C_i \to 0)_{i \in I}$$ is a filtered sequence of exact sequences, then the sequence $$0 \to \mathrm{colim}_{i \in I} A_i \to \mathrm{colim}_{i \in I} B_i \to \mathrm{colim}_{i \in I} C_i \to 0$$ is exact as well. This is part of the axiom AB5 for abelian categories.
If $\mathbf{Ab}$ was equivalent to $\mathbf{Ab}^{\mathrm{op}}$, then it would also satisfy the dual property: If $(0 \to A_i \to B_i \to C_i \to 0)_{i \in I}$ is a cofiltered sequence of exact sequences of abelian groups, then
$$0 \to \lim_{i \in I} A_i \to \lim_{i \in I} B_i \to \lim_{i \in I} C_i \to 0$$
is exact as well. But this is not the case. While it is true that the sequence is left  exact, the issue is that the map $\lim_{i \in I} B_i \to \lim_{i \in I} C_i$ does not have to be surjective.
Two examples have been described over at SE/1153414: If $n$ and $p$ are coprime, then the limit of the sequences $(0 \to \mathbb{Z} \xrightarrow{n} \mathbb{Z} \to \mathbb{Z}/n \to 0)$, where the transition maps are given by $p$, is the sequence $0 \to 0 \to 0 \to \mathbb{Z}/n \to 0$, which is not exact.
Instead, at least for $I = (\mathbb{N},\geq)$ there is a long(er) exact sequence (see lim$^1$)
$$0 \to \lim_{i \in I} A_i \to \lim_{i \in I} B_i \to \lim_{i \in I} C_i \to {\lim_{i \in I}}^1 A_i \to {\lim_{i \in I}}^1 B_i \to {\lim_{i \in I}}^1 C_i \to 0.$$
Notice that any proof of $\mathbf{Ab} \not\simeq \mathbf{Ab}^{\mathrm{op}}$ has to use infinite groups at some point. This is because for the category of finite abelian groups the functor $\mathrm{Hom}(-,\mathbb{Q}/\mathbb{Z})$ defines an equivalence of categories $\mathbf{FinAb} \simeq \mathbf{FinAb}^{\mathrm{op}}$.
Here is an alternative proof: Assume there is an equivalence $\mathbf{Ab} \simeq \mathbf{Ab}^{\mathrm{op}}$. The abelian group $\mathbb{Z}$ is a projective object of $\mathbf{Ab}$. So the image would be a projective object of $ \mathbf{Ab}^{\mathrm{op}}$, i.e. an injective object of $\mathbf{Ab}$, i.e. a divisible abelian group $T$. Since $\mathbb{Z}$ is indecomposable, $T$ is indecomposable as well. The classification of divisible abelian groups implies that either $T \cong \mathbb{Z}/p^{\infty}$ for some prime $p$ or $T \cong \mathbb{Q}$. But the equivalence also gives us an isomorphism of rings $\mathbb{Z} = \mathrm{End}(\mathbb{Z}) \cong \mathrm{End}(T)$. Since $\mathrm{End}(\mathbb{Z}/p^{\infty}) \cong \mathbb{Z}_p$ and $\mathrm{End}(\mathbb{Q}) \cong \mathbb{Q}$ we get a contradiction.
