2
$\begingroup$

Let $(Y,\tau)$ and $(X_i,\tau_i),i=1,2,...n$ be the topological spaces. Further for each i , let $f_i$ be a mapping of $(Y,\tau)$ into $(X_i,\tau_i)$. Porve that the mapping $f:(Y,\tau)\to\prod_\limits{i=1}^{n}(X_i,\tau_i)$, given by $f(y)=(f_1(y),f_2(y),...,f_n(y))$ is continuous if and only if every $f_i$ is continuous.

My proof:

$\rightarrow$ Let $U$ be an open set of $(\prod_\limits{i=1}^{n}X_i,\tau_i)$ Suppose $f$ is continuous then $f^{-1}(U)\in\tau$

$f_i(U)=p_i\circ f(U)$ where $p_i$ is the projection $p_i:(\prod_\limits{i=1}^{n}X_i,\tau_i)\to (X_i\tau_i)$.

As $f_i(U)$ is a composition of two continuous functions hence it is continuous.

$\leftarrow$ Let $U=U_1\times U_2\times...\times U_i\times...\times U_n$ be an open set in the subspace topology $f(Y),\tau_{nY}$

Supposing $f_i$ is continuous forall the $i=1,2...n$ then

$f^{-1}_i\circ p_i(U)=f^{-1}(U_i)\in\tau_i$

But $f^{-1}(U)=f^{-1}_i\circ p_i(U)$ then $f$ is continuous.

Question:

Is my proof right? If not. Why not?

Thanks in advance!

$\endgroup$
1
$\begingroup$

I would recommend the following change

$\leftarrow$ Let $U=U_1\times U_2\times...\times U_i\times...\times U_n \subset \prod_\limits{i=1}^{n}(X_i,\tau_i)$ with each $U_i$ open $X_i$, so that $U$ is open in the product space. These open sets form a basis, so we only have to show that $f^{-1}(U)$ must be open in $Y$ to prove the continuity of $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.