Is the injective map $f: X^{n-1}\to X^n$ an embedding? Let, $X$ be a CW complex , $X^{(n)}$ denotes the $n^{th}$ skeleton. We have an injective map, $f:X^{n-1}\to X^n.$
We know that $f$ is continuous, but is it a topological embedding? In other words, is the quotient topology on $X^{n-1}$ the same as the subspace topology induced as a subspace of $X^n?$
Any suggestions?
 A: If $X^{(n - 1)}$ is compact (i.e. it has finitely many cells) then the map $f : X^{(n - 1)} \to f(X^{(n - 1)})$ is a continuous bijection from a compact space to a Hausdorff space so is a homeomorphism. Otherwise, without further constraints on $f$, it's easy to construct counterexamples; for example by taking a cw structure on $X = X^{(1)} = \mathbb{R}$ with $X^{(0)} = \mathbb{Z}$.
I now see from @LeeMosher's comment that I probably misunderstood the question, $f$ is supposed to be the obvious inclusion $X^{(n - 1)} \to X^{(n)}$, not just some continuous injective map. This map is indeed a homeomorphism onto its image. It factors as $X^{(n - 1)} \hookrightarrow X^{(n - 1)} \sqcup (\bigsqcup D^n_\alpha) \to X^{(n)}$ where $\alpha$ ranges over the $n$-cells. Now to see that the composition is closed, it is enough to check that a closed subset $F \subset X^{(n - 1)} \subset X^{(n - 1)} \sqcup \bigsqcup D^n_\alpha$ has closed saturation. But the saturation of this set is $F \sqcup \bigsqcup \Phi_\alpha^{-1}(f(F))$, where $\Phi_\alpha : D^n_\alpha \to X^{(n)}$ are the characteristic maps. This is closed since each $\Phi_\alpha$ is continuous.
