# For continuous functions $f_1:X\to Y_1$ and $f_2:X\to Y_2$ prove $f_1\nabla f_2:X\to Y_1\times Y_2$ is continuous

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?

• How is the distance function defined in $Y_1\times Y_2$? – Franklin Pezzuti Dyer Dec 4 '18 at 21:21
• X, Y1, and Y2 are all topological spaces – general1597 Dec 4 '18 at 21:27
• "... 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. – zipirovich Dec 4 '18 at 22:37
• Hint: Given $U\times V\subset X_1\times X_2$ then $(f_1\nabla f_2)^{-1}(U\times V)=f_1^{-1}(U)\cap f_2^{-1}(V)$. – 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