Does the arbitrary coproduct, i.e. the arbitrary direct sum of metrizable vector spaces exist? I know already, that for metric spaces the coproduct doesn't exists, because one considers short maps as morphisms, which ultimately leads to the failure.
My own idea was to sum up chosen metrics of the $X_i$ in a suitable way: Let $\{X_i\}$ be a family of metrizable vector spaces and choose a family of metrics $\{d_i\}$ generating the topology of $X_i$. For the direct sum we take the direct sum as with vector spaces, so that we have elements of the form \begin{align} x=\sum\limits_ix_i \end{align} with $x_i\in X_i$ and only finitely many nonzero. For $x,y\in \oplus_i X_i$ we set \begin{align} d(x,y)=\sum\limits_id(x_i,y_i). \end{align}
However, I do not see, how the arising topology is independent of the choice of metrics (which should be the case, for the coproduct to be unique up to topological isomorphism). Also, I am in particularly interested in the setting of complete locally convex metrizable vector spaces, i.e. Fréchet spaces. For locally convex spaces, I think one could choose one seminorm out of every space $X_i$ and write a similar sum as above. Varying over all possible seminorms, this yields a family of seminorms on $\oplus_i X_i$. However, in the setting of Fréchet spaces, this family needs to stay countable. Could it be, that this is just not possible for Fréchet spaces, and that an arbitrary direct sum of Fréchet spaces need not be a Fréchet space?