For functions $f_1:X\to Y_1$ and $f_2:X\to Y_2$ define $f_1\nabla f_2:X\to Y_1\times Y_2$ by $(f_1\nabla f_2)(x)=(f_1(x),f_2(x))$.
Prove that $f_1\nabla f_2$ is continuous if and only if both $f_1$ and $f_2$ are continuous.
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So, $f_1\nabla f_2$ is continuous then for each $x$ in $X$, and each neighborhood $V$ of $f_1\nabla f_2$, there is a neighborhood $U$ of $x$ such that $f_1\nabla f_2(U)$ contained in $V$. Let $x_1$ be a point in $X$ with $y_1=f_1(x_1)$ and $y_2=f_2(x_1)$.
Choose neighborhood $Vy_1$ and $Vy_2$ around $y_1$ and $y_2$ respectively. Now we have to find a neighborhood $Ux_1$ around $x_1$ such that $f_1\nabla f_2(x_1)$ contains $Vy_1$ and $Vy_2$.
Is this correct so far? If so, how should I proceed? Also, what is a good way to prove the opposite direction?
You can prove this result:
If $\pi_i: Y_1\times Y_2 \to Y_i$ is the projection on the $i$-th factor then a function $g:X\to Y_1\times Y_2$ is continuous if and only if $\pi_1\circ g$ and $\pi_2\circ g$ are continuous.
In your case you have that $\pi_1\circ (f_1\nabla f_2)=f_1$ and $\pi_2\circ (f_1\nabla f_2)=f_2 $ that are continuous so $f_1\nabla f_2$ is continuous.
You can prove the results in this way:
If $g$ is continuous then $\pi_1\circ g$ and $\pi_2\circ g$ is continuos because $\pi_1$ and $\pi_2$ are continuous.
Conversely Suppose that $\pi_1\circ g$ and $\pi_2\circ g$ are continuous. Let $A=V\times W$ a basic neighborhood of $Y_1\times Y_2$.
Then $A=\pi_1^{-1}(V)\cap \pi_2^{-1}(W)$ and so
$g^{-1}(A)=g^{-1}(\pi_1^{-1}(V))\cap g^{-1}(\pi_2^{-1}(W))=$
$=(\pi_1\circ g)^{-1}(V)\cap (\pi_2\circ g)^{-1}(W) $
which is open