If a map $M$ has the feature $$ M(x+y)=M(y)M(x) $$ is there a term to describe this kind of map? It's not a linear map, at least. The order matters here - $M(y)M(x) \neq M(x)M(y)$.
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4You can't have $M(y)M(x)\neq M(x)M(y)$, because that would imply that $M(x+ y)\neq M(y + x)$, and that violates either what we want functions to be, or what we (usually) want addition to be. – Arthur Oct 13 '17 at 09:09
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But what if we have a function on the set of ordinal numbers: Then we have $\omega+1 \neq 1+\omega$, so what's the problem since each of those $2$ ordinals may have an image? – asdf May 16 '18 at 08:50
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@Matti: Can you clarify if your question is about arbitrary (possibly non-commutative) addition/multiplication or not? Or is it just about functions from $\Bbb R \to \Bbb R$? Do you assume continuity? – Martin R Nov 06 '19 at 11:47
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@MartinR I asked this question over two years ago. It's been answered already. Thank you. – Matti P. Nov 06 '19 at 11:48
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Its name is an exponential map.
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A map $M: A \to B$ with the property that $$ \forall x, y \in A: M(x+y)=M(x)M(y) \tag{*} $$ is called a group homomorphism from the (additive) group $(A, +)$ to the (multiplicative) group $(B, \cdot)$. Example: $$ M : \Bbb C \to \Bbb C, z \mapsto e^{\operatorname{Re}(z)}. $$
Your condition $$ \forall x, y \in A: M(x+y)=M(y)M(x) \tag{**} $$ is equivalent to $(*)$ if $(A, +)$ is commutative (i.e. $A$ is an abelian group), as mentioned in the comments. I am not aware of a term for $(**)$ in the non-commutative case.
In the particular case that $A=B= \Bbb R$, $(*)$ is equivalent to $f = \log M$ being an additive function on $\Bbb R$, i.e. it satisfies Cauchy's functional equation $$ \forall x, y \in \Bbb R: f(x+y)=f(x) + f(y) \, . $$ This functional equation has nonlinear solutions, but under some conditions (i.e. continuity) it follows that $f(x) = cx$ for some constant $c$. So in this special case we have that $M$ is an exponential function: $M(x) = e^{cx}$.
Martin R
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