# Surjectivity of a linear map over an infinite dimensional tensor product

I’d really some help on this problem i’ve been working on for while. I’ve got two sets $$X$$ and $$Y$$, the map $$overline{phi}:mathbb{R}^X times mathbb{R}^Y rightarrow mathbb{R}^{Xtimes Y}$$ such that $$overline{phi}(f,g)=f(x)times g(y)$$. $$overline{phi}$$ is clearly bilinear under the standar definition of function adition and scalar multiplication, so there exists a linear map $$psi :mathbb{R}^X otimes mathbb{R}^Y rightarrow mathbb{R}^{X times Y}$$ such that $$overline{phi}(f,g)=psi (fotimes g)$$.

I need to prove $$psi$$ is onto iff $$X$$ or $$Y$$ are finite.

So far i have tried defining a basis for
$$mathbb{R}^X times mathbb{R}^Y$$ as:
$$begin{equation} notag {(delta_{x_1},0),dots, (delta_{x_{|x|}},0), (0,delta_{y_1}),dots , (0, delta_{y_{|Y|}}) } end{equation}$$
Where $$delta_{x_i}(x)=1 text{if} x_i=x$$, otherwise it’s zero. Thus, a basis for the tensor product will be $${ (delta_{x_i}otimes delta_{y_j})|x_iin X, y_jin Y }$$. A basis for $$mathbb{R}^{Xtimes Y}$$ is $${ delta_{x_i,y_j}(x,y)|x_iin X, y_jin Y }$$, then i could always map $$delta_{x_i}otimes delta_{y_j}$$ into $$delta_{x_i,y_j}(x,y)$$ and get an isomorphism. $$psi$$ is one to one so it makes sense, but i can’t see how this fails if both X and Y have infinite elements.

I also tried other things like proving the surjectivity for $$overline{phi}$$, an element that can not be in the image of $$psi$$, or even showing $$(overline{phi}, mathbb{R}^{Xtimes Y})$$ didn’t have the universal property when both sets are infinite, but it didn’t seem to work for me.
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Any kind of help on this would be highly appreciated.

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