39,725 reputation
247119
bio website
location
age
visits member for 3 years, 11 months
seen 5 hours ago

Jan
14
comment Proving every subset of a finite set is finite
What's your definition of finite set? What's an infinite set? The proof will depend on the details.
Jan
14
answered Examples of Monads and their Algebras
Jan
13
comment What is the free category on the underlying graph of a category?
There is a canonical map from $D$ to the free category on $D$, but it is in general not an isomorphism or an equivalence.
Jan
13
comment What is the free category on the underlying graph of a category?
Suppose given a pair of morphisms $f, g$ with composite $h=g\circ f$. Then in the free category has morphisms $f,g,h$ as well, but now $h\ne g\circ f$.
Jan
13
accepted Cartesian closed categories with double negation elimination
Jan
13
comment What is the free category on the underlying graph of a category?
Almost always. For example, the free category on the empty graph is the empty category; but as soon as there are two composable arrows then the free category is different.
Jan
13
comment What is the free category on the underlying graph of a category?
There is a forgetful functor from the category of categories to the category of (directed multi)graphs. The free functor is the left adjoint. The explicit construction is more-or-less what you expect: it has morphisms for each generating edge, as well as all formal compositions of such.
Jan
13
comment The free abelian group monad
Hmmm, yes and no. The argument here is the usual one used to show that the category of models for an algebraic theory satisfies the split coequaliser condition of Beck's monadicity theorem, but it is not crucial in the theorem itself.
Jan
13
comment All classes of finite structures are axiomatizable in $L_{\infty\omega}$
I guess you should reduce the problem to showing that there is at most a set of isomorphism classes...
Jan
13
comment The free abelian group monad
Yes, I did assume that. But $T f$ for any function $f$ is quite easily seen to be a homomorphism of abelian groups, especially since it sends generators to generators.
Jan
13
comment Objects of the Category $\mathbf{Mat}_\mathbb K$
Sure. But using $n$ instead of $K^n$ "strongly emphasizes that objects are but labels with no internal structure."
Jan
13
comment The free abelian group monad
I added a postscript. We need to use the $T$-algebra axioms and lift to $T(T X)$ at one point.
Jan
13
revised The free abelian group monad
added 138 characters in body
Jan
13
revised The free abelian group monad
added 491 characters in body
Jan
13
comment The free abelian group monad
Sure – we can check the equations in $T (X)$ using $\eta$ and $h$.
Jan
13
answered The free abelian group monad
Jan
12
comment The free abelian group monad
OK. (I was worried you would consider it cheating.)
Jan
12
comment The free abelian group monad
I think the trick is to observe that $T (X)$ already has a canonical abelian group structure, induced by the group operation of $\mathbb{Z}$. Then define $x + y$ as $h(\eta(x) + \eta(y))$.
Jan
12
comment Passing to quotients via quotient maps preserving topological properties
The quotient topology on $Y$ is the finest topology making $p : X \to Y$ continuous; if $p$ is a quotient map then by definition $Y$ has the quotient topology, and obviously the discrete topology on $Y$ makes $p : X \to Y$ continuous if $X$ is also discrete.
Jan
12
answered Composition of a kernel with a cokernel has a normal image