Is the projection $\pi_{i_1,i_2,\ldots ,i_k}:\Pi_{i=1}^n X_i \to \Pi_{j=1}^k X_{i_j}$ always a continuous function in the product topology?

Question

Let $X_i,\; i=1,2,\ldots,n$ be topological spaces. It is well know that the canonical projections $\pi_m:\Pi_{i=1}^nX_i\to X_m$ are continuous functions for any $m\in \{1,2,\ldots,n \}$.

My question is: Is the projection $\pi_{i_1,i_2,\ldots ,i_k}:\Pi_{i=1}^n X_i \to \Pi_{j=1}^k X_{i_j}$ given with $\pi_{i_1,i_2,\ldots ,i_k}(x_1,x_2,\ldots,x_n)=(x_{i_1},x_{i_2},\ldots ,x_{i_k})$, for $i_1,i_2,\ldots ,i_k\in\{1,2,\ldots,n\}$ always a continuous function in the product topology?

I expects an affirmative answer, but I'm asking for a proof. My idea is to use the fact that canonical projections are continuous mappings and to see the new projection as an ordered pair whose coordinates are the respective canonical mappings, but I'm not sure if this guaranties the continuity of the new projection.

Answer 1

You just have to check whether $\pi^{-1}_{i_1,\dots,i_k}(U_{i_1} \times \dots \times U_{i_k})$ is open, where $U_{i_j}$ are open in $X_{i_j}$. But this set is simply the product of $X_{r}$'s and $U_{s}$'s (where the $X_r$ appear if $r \notin \{i_1,\dots, i_k\})$, and that is clearly open in the product topology since it's a product of open sets.

Answer 2

A nice lemma:

A function $f:Z\to\prod_{j=1}^n Y_i$ is continuous if and only if $\pi_j\circ f:Z\to Y_j$ is continuous for every $i\in\{1,\dots,k\}$.

Applying that on $Z=\prod_{i=1}^nX_i$ and $Y_j=X_{i_j}$ you will find immediately that your function is continuous.