If $\phi:X\times Y\rightarrow Z$ is continuous then is $\varphi(x_1,x_2,y):=\phi(x_1,y)-\phi(x_2,y)$ continuous too? So let be $X$ and $Y$ topological spaces and let be $Z$ a topological vector space. So if $\phi:X\times Y\rightarrow Z$ is a continuous function then we define a function $\varphi:X\times X\times Y\rightarrow Z$ thorugh the equation
$$
\varphi(x_1,x_2,y):=\phi(x_1,y)-\phi(x_2,y)
$$
for any $(x_1,x_2,y)\in X\times X\times Y$ and thus let we prove that it is continuous. So we observe that the statement follows directely proving that the function $\Delta:X\times X\times Y\rightarrow Z\times Z$ defined through the equation
$$
\Delta(x_1,x_2,y):=\big(\phi(x_1,y),\phi(x_2,y)\big)
$$
for any $(x_1,x_2,y)\in X\times X\times Y$ is continuous since $\varphi$ would be composition of continuous functions. So to prove hte continuity of $\Delta$ I tried to implement the universal mapping theorem for products but unfortunately I did no able to conclude anything. So could someone help me, please?
 A: The function $\Delta : (x_1,x_2,y) \mapsto (\phi(x_1,y),\phi(x_2,y))$ is continuous iff the two functions $f:(x_1,x_2,y) \mapsto \phi(x_1,y)$ and $g:(x_1,x_2,y) \mapsto \phi(x_2,y)$ are continuous.
To see that this is the case, notice that $f$ is the composition of $\phi$ and $(x_1,x_2,y) \mapsto (x_1,y)$ and that both are continuous.
A: So let be $p_i:X\times X\times Y\rightarrow X\times Y$ for $i=1,2$ the function defined through the equation
$$
p_i(x_1,x_2,y):=(x_i,y)
$$
for any $(x_1,x_2,y)\in X\times X\times Y$ and the let we prove that it is continuous. So if $U\times V$ is a basic open set of $X\times Y$ then we observe that the sets
$$
p^{-1}_1[U\times V]=U\times X\times Y\,\,\,\text{and}\,\,\,p^{-1}_2[U\times V]=X\times U\times V
$$
are open so that $p_i$ for $i=1,2$ is continuous. So if $\pi_i$ is for $i=1,2$ the $i$-th projection of $Z\times Z$ in $Z$ then we observe that
$$
\pi_i\circ\Delta=\phi\circ p_i
$$
for each $i=1,2$ so that we conclude that the function $\Delta$ is continuous. Now we know that the diagonal product of continuous functions is continuous too (see here for details) so that if $\Lambda$ is diagonal product of the identity map over $Z$ and of the restriction to $Z\times\{-1\}$ of the product beetween scalars and vectors then $\Lambda$ is continuous. So if $\sigma$ is the sum between vectors of $Z$ then we observe that
$$
\varphi(x_1,x_2,y)=\sigma\big(\phi(x_1,y),-\phi(x_2,y)\big)=\sigma\Big(\Lambda\big(\phi(x_1,y),\phi(x_2,y)\big)\Big)=\\
\sigma\Biggl(\Big(\Lambda\big(\Delta(x_1,x_2,y)\big)\Big)\Biggl)=\big(\sigma\circ\Lambda\circ\Delta\big)(x_1,x_2,y)
$$
for each $(x_1,x_2,y)\in X\times X\times Y$ so that we conclude that $\varphi$ is continuous.
A: Continuity of $\Delta$ (when $Z \times Z$ has the product topolgy) is equivalent to continuity of the functions $(x_1,x_2,y) \mapsto \phi (x_1,y)$ and $(x_1,x_2,y) \mapsto \phi (x_2,y)$. The inverse image of an open set $V$ in $Z$ under the first of these is of the form  $\{(x_2,x_2,y): (x_1,y) \in T, x_1 \in Y\}$  where $T$ is open in $X \times Y$ and this set is open. Similarly for the second map.
