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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.

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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 $f_2$.

Now it only remains to prove a

Lemma: $c(Y_1) \otimes c(Y_2)$ is canonically isomorphic to $c(Y_1 \otimes Y_2)$ as $\Bbb C$-vector spaces.

The proof of the lemma is a simple exercise, and the original question immediately follows from the lemma.

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