1
$\begingroup$

Suppose we're given an associative operation $\star : X \times X \rightarrow X$. Then for each $n \in \mathbb{N}_{>0}$, there's a function $f_n : X^n \rightarrow X$ given as follows:

$$f_n(x_0,\ldots,x_{n-1}) = x_0 \star \cdots \star x_{n-1}$$

Taken as a whole, the family $f_*$ has:

  • the left extension property:

$$f_a(\tilde{x}) = f_b(\tilde{y}) \rightarrow f_{a+1}(z,\tilde{x}) = f_{b+1}(z,\tilde{y})$$

  • and the right extension property

$$f_a(\tilde{x}) = f_b(\tilde{y}) \rightarrow f_{a+1}(\tilde{x},z) = f_{b+1}(\tilde{y},z)$$

which hold for all $a,b \in \mathbb{N}_{>0}$ and all $\tilde{x} \in X^a$ and $\tilde{y} \in Y^b$.

If we weaken these conditions so that they're only required to hold when $a=b$, then its easy to find examples of families $f_*$ that aren't induced by an associative operation. For example, the "averaging" family $$f_n(x_0,\ldots,x_{n-1}) = \frac{x_0+\cdots+x_{n-1}}{n}$$ satisfies the "weak" version of each extension property.

Question. Is every family $f_*$ satisfying both left and right extension properties induced by an associative operation?

$\endgroup$
1
$\begingroup$

Let $X=\mathbb{N}$ and define $f_n:X^n\to X$ by $f_n(\tilde{x})=n$ for all $\tilde{x}$. Then this trivially satisfies your extension properties, but cannot come from an associative operation.

On the other hand, if you require $f_1$ to be the identity map, then every such family does come from an associative operation. Indeed, for any $x,y,z\in X$, we have $f_2(x,y)=f_1(f_2(x,y))$ and so by right extension $f_3(x,y,z)=f_2(f_2(x,y),z)$. Similarly, $f_3(x,y,z)=f_2(x,f_2(y,z))$. This says $f_2$ is associative and $f_3$ is the induced ternary operation, and similar arguments show $f_n$ is the induced $n$-ary operation for each $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.