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.


Is my proof right? If not. Why not?

Thanks in advance!


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 $f$.


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