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3

So to clarify, this is true in a spherical category (when you assume that left and right traces are equal, this is the definition of a spherical category). The best way (IMO) to prove this is through graphical calculus. Here is a picture (which I hope is not too horrible). The only non-obvious equality is the first: it follows from the monoidality of the ...

2

Yes, it means the hom spaces have all been tensored with $\overline{R}$. (Note that this notation is consistent with the case that $C$ has one object, in which case it is a commutative ring.)

1

$\require{AMScd}$ Pick natural transformations $\alpha:F_1\to F_2$ and $\beta:G_1\to G_2$. For any $X$ consider $$\begin{CD} F_1G_1X @>\alpha_{G_1X}>> F_2G_1X \\@VV{F_1\beta_X}V @VV{F_2\beta_X}V\\ F_1G_2X @>{\alpha_{G_2X}}>> F_2G_2X \end{CD}$$ Naturality of $\alpha$ applied to the vertical maps $\beta_X:G_1X\to G_2X$ implies this is ...

3

So, you know that if you have a monoid $m$ in a monoidal category $(M, \otimes)$ then it's a natural thing to do to look at actions of $m$, namely morphisms $m \otimes c \to c$ satisfying etc. where $c$ is another object in $M$. One way to motivate the definition of monads is that you can actually do something more general than this: $c$ need not be an ...

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An object $X$ of an abelian category has finite length if it has a (finite) composition series: i.e., a chain of subobjects $$0=X_0<X_1<\dots<X_{n-1}<X_n=X$$ such that $X_i/X_{i-1}$ is simple for $1\leq i\leq n$.

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