Let $X$ and $Y$ be topological spaces, and for any $x\in X$, let $f: Y\rightarrow X\times Y$ be $f(y)=(x_0,y)$ for $y\in Y$. How do I show that $f$ is an embedding?
I am using the definition that a continuous injection that is a homeomorphism onto its image is an embedding. To prove this do I need to show 1. continuous 2. injective and 3. homeomorphism? I am a bit confused, since the definition of homeomorphism requires continuous for both directions, and also bijectivity.
My work so far: 1. continuous: if a set of $(x,y)$ is open in $X\times Y$, then the $x,y$ must be open in $X$ and $Y$. (but why exactly?)
2. injectivity: quite clear
3. homeomorphism: how is this not trivial? Can someone explain this please?
Thanks
Best Answer
A function $f:X\to\prod_{i\in I} Y_i$, where $\prod_{i\in I} Y_i$ has the product topology, is continuous iff $\pi_i\circ f$ is continuous for all $i\in I$, where $\pi_i:\prod_{i\in I} Y_i\to Y_i$ is the $i$-th projection.
In your problem $\pi_1\circ f$ is a constant function and $\pi_2\circ f$ is the identity, therefore $f$ is continuous.
Note that its image is $\{x_0\}\times Y$, so you need to prove $f:Y\to\{x_0\}\times Y$ is a homeomorphism. We already know it is continuous and bijective. It remains to show that $f$ is open onto $\{x_0\}\times Y$ (or equivalently $f^{-1}$ is continuous). This is quite obvious because if $U\subseteq Y$ is open then $f(U)=\{x_0\}\times U$ is open in $\{x_0\}\times Y$.