I am self-studying Lee's "Introduction to Topological Manifolds" and am learning about the product topology in Chapter 3. There is a large proposition containing several properties of the product topology, and I'm trying to prove each of these.
Following is the part (c) that I'm working on; would someone critique my approach? I'm thinking it seems over-complicated.
Proposition 3.12 Let $X_1, \ldots, X_n$ be topological spaces. (c) For any $i$ and any points $x_j \in X_j$, $j\neq i$, the map $f_i: X_i \rightarrow X_1 \times \cdots \times X_n$ given by $$ f_i\left(x\right) = \left(x_1, \ldots, x_{i-1}, x, x_{i+1}, \ldots, x_n\right)$$ is a topological embedding of $X_i$ into the product space.
For completeness, the book defines a topological embedding of topological space $A$ into $B$ as an injective, continuous mapping $f:A\rightarrow B$ which is a homeomorphism onto $f\left(B\right)$. I've interpreted this more simply as a homeomorphism $f:A \rightarrow f\left(A\right) \subseteq B$, where $f\left(A\right)$ has the subspace topology from $B$.
Directly proceeding this proposition is the statement (and proof) of the following characteristic property of the product topology:
Theorem (Characteristic Property of Product Topologies) Let $X_1 \times \cdots \times X_n$ be a product space. For any topological space $B$, a map $f: B \rightarrow X_1 \times \cdots \times X_n$ is continuous if and only if each of its component functions $f_i = \pi_i \circ f$ is continuous.
My Attempt
Proof Let $X = X_1 \times \cdots \times X_n$ be a product of topological spaces, and let $f_i:X_i \rightarrow X$ be given as in the proposition statement. Since $X$ is a product topological space, it satisfies the characteristic property above; applying it to $f_i$, we have $$\begin{align} f_i: X_i\rightarrow X \text{ is continuous }& \iff \pi_i \circ f_i \text{ is continuous} \end{align}$$ But $\pi_i \circ f_i = \text{id}_{X_i}$ which is known to be continuous. Thus, $f_i$ is continuous.
To show that $f_i$ is a topological embedding of $X_i$ into $X$, I need to show that it is a homeomorphism onto its image, namely $$ f_i\left(X_i\right) = \left\{x_1\right\} \times \cdots \times X_i \times \cdots \times \left\{x_n\right\}.$$
It's easy to show that $f_i$ is injective; it is trivially surjective onto its image. We showed above that $f_i: X_i \rightarrow X$ is continuous, and from properties of subspace topologies, it follows that $f_i: X_i \rightarrow f_i\left(X_i\right)$ is also continuous. It remains to show that $f_i$ has a continuous inverse. I will do this by showing that $f_i$ is an open map.
Let $U \subset X_i$ be open. Then $$ f_i\left(U\right) = \left\{x_1\right\} \times \cdots \times U \times \cdots \times \left\{x_n\right\} = \left(X_1 \times \cdots \times U \times \cdots \times X_n \right) \cap f_i\left(X_i\right). $$ Note that the left operand of this intersection is open in the product topology on $X$, and so the intersection shows that $f_i\left(U\right)$ is open in the subspace topology on $f_i\left(X_i\right)$. Thus, I have shown that $U\subset X_i$ open implies that $f_i\left(U\right) = \left(f_i^{-1}\right)^{-1}\left(U\right)$ is open in $f_i\left(X_i\right)$, so that $f_i^{-1}$ is continuous. Putting this all together, we have $f_i:X_i \rightarrow f_i\left(X_i\right)\subset X$ is a homeomorphism, so $f_i$ is a topological embedding of $X_i$ into $X$. $\blacksquare$
Is this correct, and is there an easier way to prove this, perhaps using that characteristic property? While playing around I noticed that $f_i^{-1}: f\left(X_i\right) \rightarrow X_i$ is exactly a restriction of the projection map: $$f_i^{-1} = \pi_i \vert_{f\left(X_i\right)}. $$ While $\pi_i$ is continuous and we know that a continuous mapping restricted to an open subset of its domain is still continuous, I don't think I can apply this to say that $f_i^{-1}$ is continuous since $f\left(X_i\right)$ doesn't seem open in $X$.
One has to show, that $f_i : X_i \to f_i(X_i)$ is a homeomorphism. Clearly it is continuous by definition resp. characteristic property of subspace topology.
It is by definition of subspace topology $\iota: f_i(X_i) \hookrightarrow X_1\times ... \times X_n$ continuous.
Let $g := \pi_i\circ \iota$, i.e. $g = \pi_i|_{f_i(X_i)}$. Then $g$ is continuous as a composition of continuous maps.
$f_i \circ g = id_{f_i(X_i)}$ and $g\circ f_i = id_{X_i}$.
So, $f_i$ is a homeomorphism by definition of homeomorphism.