map of endomorphism operad Let $\mathbf{Top}$ be the category of (nice) topological spaces.
For any space $Z$, define $\mathbf{End}_{\text{operad}}(Z)$ as the endomorphism operad.
Is there always a map of operads
$$\mathbf{End}_{\text{operad}}(Z\times Z)\rightarrow \mathbf{End}_{\text{operad}}(Z) $$
Edit: Intuitively, I would say the answer should be NO i.e. there is NOT ALWAYS such a map of operads. But I don't have a concrete counterexample.
 A: The answer is no already for $Z = \mathbb{R}$. The endomorphism operad of $\mathbb{R}^2 \cong \mathbb{C}$ contains the Lawvere theory of $\mathbb{C}$-algebras, which cannot act on $\mathbb{R}$. This amounts to proving the following slightly strange statement:

No topological $\mathbb{C}$-algebra (meaning, for our purposes a $\mathbb{C}$-algebra equipped with a topology such that addition, multiplication, and scalar multiplication by a fixed $\lambda \in \mathbb{C}$ are all continuous; importantly, we don't require continuity in $\lambda$) can be homeomorphic to $\mathbb{R}$.

Proof. Suppose $A$ is a topological ring homeomorphic to $\mathbb{R}$. Then $(A, +)$ is a topological group homeomorphic to $\mathbb{R}$. By Gleason-Montgomery-Zippin's solution to Hilbert's fifth problem, $(A, +)$ is a $1$-dimensional simply connected Lie group, hence $(A, +) \cong (\mathbb{R}, +)$ as topological groups; from now on we will write $A = \mathbb{R}$.
(Edit: This is overkill, as expected. There is a totally elementary argument that a topological group (not necessarily abelian!) homeomorphic to $\mathbb{R}$ must be isomorphic to $(\mathbb{R}, +)$; see this math.SE question.)
Write $\cdot : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ for the multiplication on $A$. By rescaling we may assume WLOG that the multiplicative identity is $1$. It follows that $q_1 \cdot q_2 = q_1 q_2$ for rational $q_1, q_2$, and then by continuity it follows that $r_1 \cdot r_2 = r_1 r_2$ for all $r_1, r_2$, so $A \cong \mathbb{R}$ as a topological ring.
(So $(\mathbb{R}, +, \times)$ is the unique topological ring homeomorphic to $\mathbb{R}$, which is a fun fact that I don't think I've seen before.)
In particular, $A \cong \mathbb{R}$ as a ring, so admits no $\mathbb{C}$-algebra structure (even one where scalar multiplication may be discontinuous) because there is no ring homomorphism $\mathbb{C} \to \mathbb{R}$ (since $\mathbb{R}$ does not have an element satisfying $x^2 = -1$). $\Box$
Here I've blithely ignored the difference between a morphism of Lawvere theories and a morphism of operads, but I don't think it matters: concretely, the endomorphism operad of $\mathbb{R}^2 \cong \mathbb{C}$ contains binary operations describing addition and complex multiplication, and unary operations describing complex scalar multiplication, and nullary (or unary depending on taste) operations describing the additive and multiplicative identities, satisfying the $\mathbb{C}$-algebra axioms, and a morphism into another endomorphism operad $\textbf{End}(X)$ implies equipping $X$ with the same operations satisfying the same axioms, and that's all we need (even if there may or may not be other weird things going on involving the diagonal and projection maps).
