Statement
Let be $W$ a topological vector space and $\phi:U\rightarrow W$ and $\psi:V\rightarrow W$ two continuous functions. So if we define $f:U\times V\rightarrow W$ though the condition $$ 1.\quad f(u,v):=\phi(u)+_{_{W}}(-1)*_{_{W}}\psi(v) $$ for any $u\in U$ and for any $v\in V$ then $f$ is continuous in the product topology.
Unfortunately I can't prove the statement: I have proved to show that $f$ is composition of continuous functions defining the function $\Delta:U\times V\rightarrow W\times W$ through the condition $$ \Delta(u,w):=\big(\phi(u),\psi(v)\big) $$ that for the universal mapping theorem for products is continuous (is this correct?) but then I can't continue because, although I see that $f(u,v)=+_{_{W}}\Big(\phi(u),*_{_{W}}\big(-1,\psi(v)\big)\Big)$, I can't prove that the function $\tilde\Delta:U\times V\rightarrow W\times W$ defined through the condition $$ \tilde\Delta(u,v)=\Big(\phi(u),*_{_{W}}\big(-1,\psi(v)\big)\Big) $$ for any $u\in U$ and $v\in V$ is continuous. Naturally $+_{_{W}}$ is the vectorial sum in $W$ and $*_{_{W}}$ is the scalar multiplication in $W$. So could someone help me, please?
We can write $f$ as the composition of simpler maps in the form $f = +_W \circ s \circ \Delta$, where the addition $+_W$ is continuous by the definition of topological vector spaces, the map $\Delta \colon (u,v) \mapsto (\phi(u),\psi(v))$ is continuous by general properties of maps between product spaces, and $s \colon (w_1, w_2) \mapsto (w_1, -w_2)$ remains to be seen as continuous.
Again by general properties of maps between product spaces, the continuity of $s$ is equivalent to the continuity of the negation map $n \colon W \to W$, $n(w) = -w$. We can write this as the composition $$w \mapsto (-1,w) \mapsto (-1)\ast_W w = -w$$ of the embedding $W \to \{-1\}\times W \subset K \times W$ and the scalar multiplication $K \times W \to W$. The embedding is continuous since each component is, and the scalar multiplication is continuous by definition of a TVS. Hence $n$ is continuous, and thus $s$ is continuous, and finally the continuity of $f$ follows.