Let $(Y,\tau)$ and $(X_{i},\tau_{i}),\; i=1,2,...,n\;$ be topological space. Further for each $i$, let $f_{i}$ be a mapping of $(Y,\tau)$ into $(X_{i}, \tau_{i}).$ Prove that the mapping $f:(Y, \tau) \rightarrow \Pi_{i=1}^{n}(X_{i},\tau_i)$, given by $$f(y)=<f_1(y),f_2(y),...,f_n(y)>$$ is continuous if and only if every $f_i$ is continuous
Attempt:
Consider $f_i:P_i\circ f:(Y,\tau) \rightarrow (X_i,\tau_i)$ where $P_i$ is projection mapping.
$(\rightarrow)$ Suppose $f$ is continuous then each $f_i$ is continuous since $f_i=P_i\circ f$ and $P_i$ is continuous. We know composition of two continuous functions is continuous. We are done.
$(\leftarrow)$ Suppose each $f_i$ is continuous that means for each $\mathrm{U}$ in $(X_i,\tau_i)$ we have $f_i^{-1}(\mathrm{U})$ is $(Y,\tau)$. Means also that $P_i \circ f$ is continuous. Let $\mathrm{U}\in \Pi(X_i,\tau_i)$ be open set we want to show $f^{-1}(\mathrm{U})$ is open in $(Y,\tau)$, observe that:
$$f^{-1}(\mathrm{U})=<f_1^{-1}(\mathrm{U}),f_2^{-1}(\mathrm{U}),...,f_n^{-1}(\mathrm{U})>$$
Since each $f_i$ is continuous thus $f^{-1}(\mathrm{U})$ is continuous.
Second part is not correct. $f_1^{-1}(U)$ does not make sense since $U$ is not a subset of $X_1$.
A basic open set in the product space is of the type $U=\prod U_i$ where $U_i$ is open in $X_i$ for each $i$ and $U_i=X_i$ for all but finitely many $i$. For such a set $U$ we have $f^{-1}(U)=\bigcap_i f_i^{-1}(U_i)$ which is open (because of the condition $U_i=X_i$ for all but finitely many $i$). This finishes the proof.