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If we require an isomorphism (as opposed to an equivalence) the answer is no, for reasons having little to do with the (interesting aspect of) the question. Assume there exists an abelian category $T …

Let me start with a remark [EDITED for clarity after Andrew's comments]. Given $h:X\to Y$, the following are equivalent: (1) $h$ is the coequalizer of some $W\rightrightarrows X$, (2) $h$ is the coequ …

Let me show that if $\Gamma$ preserves filtered colimits, then $X$ is quasicompact. (At the moment I don't know about 'quasiseparated'; but, as Martin points out, I only use the injectivity of $\varin …

A scheme $X$ is reduced if and only if the natural map $$  \coprod_{x\in X}\operatorname{Spec}  \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.   [EDI …

Not quite an answer, but too long for a comment. Combining the two (open/closed) gluing constructions, we can conclude that if $Z$ is an open subscheme of a closed subscheme $W$ of $X$, then it is an …

If I understand the question correctly, the category of schemes is regularly wellpowered. A locally closed immersion factors as $Y\xrightarrow{i}U\xrightarrow{j}X$ where $j$ (resp. $i$) is an open (r …

Now I am told that the category of affine schemes is wellpowered (equivalently, the category of commutative rings is cowellpowered). Let me deduce from this that the category of schemes is wellpowered …

How about the category of sets with injective maps as morphisms? As an ad hoc example, this may not count as "natural", but it's simple enough. [EDIT] following Martin's comment: take the dual, or r …

Here is a counterexample. Put $F(X)=\mathbb{Q}$ for all $X$, and for every morphism $f:X\to Y$ put $F(f)=$ multiplication by $\deg(f)$. Now assume $V$ is a $\mathbb{Q}$-vector space and $\phi:\underli …

The answer is no to all questions except 4. Negative answers to 1,2 and 3: It is easy to construct a ring $A$ with an element $a$ satisfying: (i) $a≠0$, (ii) $a$ is nilpotent, (iii) for each $n≥1 …

Answer to 2: If $A$, $B$, $C$ are three sets, it is not true in general that $(A\times B)\coprod C=(A\coprod C)\times (B\coprod C)$. Hence the category $(Sets)^{op}$ is a counterexample. EDIT: (March …

Let me show that every such $Z$ has finite length. First, note that being of the form $\varinjlim X_i$ (with the $X_i$ of finite length) is the same as being countably generated over $A$. So let us st …