bio  website  math.uwo.ca/~dschaepp 

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visits  member for  4 years, 8 months 
seen  Jul 2 at 0:02  
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I am a postdoc at the University of Western Ontario.
2d

awarded  Nice Answer 
May 12 
awarded  Enlightened 
May 12 
awarded  Nice Answer 
Apr 19 
comment 
Universal property of module categories over monads
Is it clear that $\tilde{G}$ preserves coproducts? If the categories or the monad are not additive, then coproducts in the category of modules (algebras) are usually quite different from coproducts in the base category (e.g. groups vs. sets). If $T$ is additive and finitary (preserves filtered colimits), this is certainly not a problem. 
Feb 1 
revised 
Semiadditivity and dualizability of 2
added 7 characters in body 
Feb 1 
revised 
Semiadditivity and dualizability of 2
updated expired link 
Jan 25 
comment 
Pushouts in the category of adjunctions
On the other hand, if you consider the 2category of say lfp categories and adjunctions whose right adjoint does preserve filtered colimits, then you do get the desired pushouts. This 2category is equivalent to finitely cocomplete categories with functors preserving finite colimits, and this is 2monadic over the category of small categories. Since the 2monad in question has rank, this 2category has all bicolimits. I think it would be interesting to see if there are counterexamples if the categories involved do not have (many) colimits. 
Jan 25 
comment 
Pushouts in the category of adjunctions
I don't think this works quite as intended. First of all, you often talk about limits when you really mean colimits (both in the definition of $X_i$ and $X_{\infty}$). But right adjoints need not preserve colimits (not even filtered colimits in general), so your argument that $G_1 X_{\infty}=G_2 Y_{\infty}$ does not work. 
Jan 25 
answered  Does trace handle composition in a traced symmetric monoidal category? 
Jan 25 
awarded  ct.categorytheory 
Jan 24 
answered  Pushouts of fully faithful (enriched) functors 
Jan 7 
comment 
Is a composite of (co)monadic adjunctions (co)monadic?
Sorry, the functor in question sends a graph to the disjoint union of its set of arrows and its set of vertices, not just its set of arrows. The two nontrivial elements $s,t$ of the monoid act trivially on the set of objects, and send arrows to their source and target respectively. The usual relations that source of the target is target etc. have to be imposed. 
Jan 7 
awarded  Commentator 
Jan 7 
comment 
Is a composite of (co)monadic adjunctions (co)monadic?
Well, it is also monadic over $\mathbf{Set}\times \mathbf{Set}$, but the functor which sends a graph to its set of arrows is also monadic. In fact, there is a three element monoid in $\mathbf{Set}$ whose category of actions is equivalent to graphs (see exercise (GRMN) on page 107 of "TTT"). 
Jan 7 
answered  Is a composite of (co)monadic adjunctions (co)monadic? 
Dec 6 
revised 
is the category of coherent sheaves some kind of abelian envelope of the category of vector bundles?
Added paragraph about algebraic stacks 
Dec 3 
awarded  Nice Answer 
Nov 28 
answered  is the category of coherent sheaves some kind of abelian envelope of the category of vector bundles? 
Nov 10 
awarded  Yearling 
Sep 5 
comment 
Do unbounded chain complexes have enough injectives?
@Martin: Good question! The argument I gave below works for chain complexes with values in any Grothendieck abelian category, since they are closed under the formation of diagram categories. 