# Tensor product of antilinear maps.

Let $$X_1, X_2, Y_1, Y_2$$ be $$\mathbb{C}$$-vector spaces and $$f_1: X_1 \to Y_1$$ and $$f_2: X_2 \to Y_2$$ conjugate-linear maps, i.e. $$f_1(\alpha x_1) = \overline{\alpha} f_1(x_1)$$ and $$f_1(x_1 + x_1') = f_1(x_1) + f_1(x_1')$$ and similarly for $$f_2$$.

Question: Does there exist a unique conjugate linear map $$f_1 \otimes f_2: X_1 \otimes X_2 \to Y_1 \otimes Y_2$$ such that $$(f_1 \otimes f_2)(x_1 \otimes x_2) = f_1(x_1) \otimes f_2(x_2)?$$

Basically, we would want to apply the universal property of the tensor product to construct this map, but the maps are not linear. But I believe there should be a trick to reduce the problem to linear maps.

An alternative way of approaching this: let $$\{e_i\}_{i \in I}$$ be a basis for $$X_1$$ and define $$(f_1\otimes f_2) (\sum_{i \in I} e_i \otimes z_i) := \sum_{i \in I} f_1(e_i) \otimes f_2(z_i)$$ and this gives existence of the map, but it still looks like an unnatural way to proceed.

If $$V$$ is a $$\Bbb C$$-vector space, then we can define a conjugate vector space $$c(V)$$ as follows.
The underlying abelian group of $$c(V)$$ is the same as the abelian group $$V$$. The scalar product $$\bullet$$ in $$c(V)$$ is defined as $$\alpha \bullet v = \overline\alpha \cdot v$$, where $$\cdot$$ denotes the scalar product on $$V$$.
Thus by definition, the map $$f_1$$ is nothing but a $$\Bbb C$$-linear map from $$X_1$$ to $$c(Y_1)$$. Same for $$f_2$$.
Lemma: $$c(Y_1) \otimes c(Y_2)$$ is canonically isomorphic to $$c(Y_1 \otimes Y_2)$$ as $$\Bbb C$$-vector spaces.