Suppose that $X$ is a topological space which is path-connected. Let $x_0∈ X$. Show that the function $ t_{x_0} : {X} \to {X×X}$ where $t_{x_0}(x)=(x,x_0)$ is an embedding and image of $X$ under $t_{x_0}$ is closed in ${X×X}$.
I know it is trivial that $X$ and its image are homeomorphic. My question is why $X$ need to be path-connected in order to show what the question requires. Is this related to the image of $X$ being closed under $t_{x_0}$?
I really appreciate if someone could explain why path-connectivity is required!
The projection map $p_2\colon X\times X\rightarrow X,(x_1,x_2)\mapsto x_2$ is continuous by definition of the product topology, therefore preimages of closed subsets are closed. If for $x_0\in X$ the singelton subset $\{x_0\}\subset X$ is closed, then $t_{x_0}(X)=X\times\{x_0\}=p_2^{-1}(\{x_0\})\subset X\times X$ is closed. As you can see, the condition of path-connectedness is irrelevant, only that $\{x_0\}$ is closed. The alternative condition of $X$ being a $T_1$ space would guarantee this.