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?

  • $\begingroup$ How is the distance function defined in $Y_1\times Y_2$? $\endgroup$ – Franklin Pezzuti Dyer Dec 4 '18 at 21:21
  • $\begingroup$ X, Y1, and Y2 are all topological spaces $\endgroup$ – general1597 Dec 4 '18 at 21:27
  • $\begingroup$ "... and each neighborhood $V$ of $f_1\nabla f_2$ ... " doesn't make sense -- $f_1\nabla f_2$ is a function, not one of the spaces. $\endgroup$ – zipirovich Dec 4 '18 at 22:37
  • $\begingroup$ Hint: Given $U\times V\subset X_1\times X_2$ then $(f_1\nabla f_2)^{-1}(U\times V)=f_1^{-1}(U)\cap f_2^{-1}(V)$. $\endgroup$ – PtF Dec 4 '18 at 23:04

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 continuos if and only if $\pi_1\circ g$ and $\pi_2\circ g$ are continuos.

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 continuos so $f_1\nabla f_2$ is continuos.

You can prove the results in this way:

If $g$ is continuos then $\pi_1\circ g$ and $\pi_2\circ g$ is continuos because $\pi_1$ and $\pi_2$ are continuos.

Conversely Suppose that $\pi_1\circ g$ and $\pi_2\circ g$ are continuos. 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) $

That it is open


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