# Tag Info

Accepted

### Congruences that aren't "finite from above," take 2: semigroups

This one is also trivial. Sorry, just take a semigroup with the identity xy=x. Then all equivalence relations are congruences so this is the same as pure sets. Here is a more interesting construction ...
Accepted

### When is the cofibrant replacement of a product the product of the cofibrant replacements?

I asked this question a very long time ago, when I was just starting to do research in abstract homotopy theory. This is a classic case of an xy problem, where I had a proof in mind, and by asking ...
Accepted

### Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$

Let $k_n$ be the least integer $k$ such that, for any digraph $D$ of order $n$ and any vertices $x,y\in D$, if there are $x$-$y$ walks of length $1,\dots,k$, then there are $x$-$y$ walks of all ...

### Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$

Here's an example of size 9 inspired by domotorp's anwer, where I only consider the primes $2,5$, and where I replace the additional long arrows from the basis of the cycles with arrows from the ...

### Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$

domotorp's lovely solution is by far the best one, but here is an explicit counterexample for $n = 10$, I wonder if it's computationally tractable to figure out the max $n$ for which your statement ...
### Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$
This is false as shown by the following digraph. From $x$ there is an edge to $v_p$, from $v_p$ there is a cycle of length $p$ to itself, and from $v_p$ there are $p-1$ different paths to $y$, of ...