Why is this imbedding continuous?

Question

Define the imbedding $i_{x_{1}}: Y \rightarrow X \times Y$ by $i_{x_{1}}(y) = (x_1,y)$, where $X$ and $Y$ are topological spaces. Then supposedly $i_{x_{1}}$ is continuous, but I fail to see why; we need to show that $i_{x_{1}}^{-1}(O)$ is open, where $O$ is any open subset of $X \times Y$. I don't see why this is the case; can anyone provide me a proof of why the imbedding is continuous? Thanks in advance for your time and efforts.

Answer 1

A function $f : X \to Y \times Z$ is continuous iff $\pi_Y \circ f$ and $\pi_Z \circ f$ are both continuous. (univeral property of continuity of maps into products).

In our case: $\pi_Y\circ i_{x_1} = \mathrm{id}_Y$ (the identity) and $\pi_X \circ i_{x_1} = c_{x_1}$ (a constant map) and both identities and constant maps are always continuous. Hence so is $i_{x_1}$.

Answer 2

One coordinate function is constant, the other is the identity. Both continuous. Since the product topology is the coarsest which makes the projections onto the coordinates continuous, we have the universal property @Henno refers to, and the map is continuous.