Is $\prod_{i \in I} f_i$ continuous?

Question

Let $I \neq \emptyset, (X_i \neq \emptyset, \mathcal{T}_i)_{i \in I}$ be a family topological spaces.. Choose (using AC) an element $(x_i)_{i \in I} \in \prod_{i \in I} X_i \neq \emptyset$

Define for a fixed $k \in I$ the function $f_k: X_k \to X_k: z \mapsto z$ and for $i \in I \setminus\{k\}$ the function $f_i: X_k \to X_i: z \mapsto x_i$

I'm following a proof, and I don't understand a step. The step will be solved once I can prove that

$$\prod_{i \in I} f_i: (X_k,\mathcal{T}_k) \to \left(\prod_{i \in I}X_i, \prod_{i \in I}\mathcal{T}_i\right): y \mapsto (f_i(y))_{i \in I} $$

is continuous (but it is possible that this is wrong).

Since $\{\operatorname{pr}_i(B_i) \mid i \in I, B_i \in \mathcal{T}_i\}$ is a subbasis for the product topology, it suffices to show that $$\left(\prod_{i \in I}f_i \right)^{-1}(B_i) \in \mathcal{T_k}$$ for $B_i \in\mathcal{T_i}$, but I failed to prove this.

Is the statement true? And if so, why?

Answer 1

Let $\pi_j:\prod X_i\to X_j$ be the canonical projection. By the universal property of the product topology your $\prod f_i$ function is continuous if and only if $\pi_j\circ \prod f_i$ is continuous for each $j$.

But note that $\pi_j\circ \prod f_i=f_j$ is obviously continuous.