0
$\begingroup$

I'm reading Algebraic Topology by Hatcher and there's a small passage that I don't understand. It's on page 519: "each cell $e_n^\alpha$ has its characteristic map $\Phi_\alpha$, which is by definition the composition $D_\alpha^n \hookrightarrow X^{n-1} \sqcup_\alpha D_\alpha^n \rightarrow X^n \hookrightarrow X$. (...) The restriction of $\Phi_\alpha$ to the interior of $D_\alpha^n$ is a homeomorphism onto $e_\alpha^n$."

I had problems when the dimension of X is infinity. I can understand why the restriction of $\Phi_\alpha$ defines a homeomorphism into the image in $X^n$. We use just properties of attaching spaces. But when we consider the weak topology in $X$, why the inclusion $i_n : X_n \rightarrow X$ preserves this property?

Also, he writes $X^n \hookrightarrow X$. For me, the symbol "$\hookrightarrow$" means that $X_n$ is homeomorphic to $i_n(X_n)$ in $X$. Is it correct? My second doubt is about his homeomorphism: I've studied topological vector spaces and there are some conditions to ensure that when we have $X_1 \subset X_2 \subset \dots X$, $X_i$ with the topology of subspace of $X_{i+1}$ and we define the weak topology in $X$, then the topology induced in $X_i$ by $X$ is the 'original' topology on $X_i$. If it's true in this case, then I can understand why he wrote $X^n \hookrightarrow X$. Is it true or is it a consequence of another fact?

$\endgroup$
1
$\begingroup$

What the arrow "$\hookrightarrow$" means is basically that it is injective(not always a homeomorphism onto its image! this would be the definition of an embedding. even though in this case it is). here it even means inclusion, therefore a homeomorphism onto its image.

So actually this already answers your question.

But one note on why inclusions, continuity or homeomorphisms are preserved under the limit. You just have to check, that those conditions are satisfied for every finite complex $X^n$. Because then the topology of $X^\infty$ respects such properties by definition: $U \subset X^\infty$ is open if and only if $U\subset X^n$ is open for all $n$.

So since everything is made to be compatible, e.g. the inclusions $X^n\hookrightarrow X^{n+1}$ are continous, if you regard a subspace contained in a finite complex, with subspace topology, it will always coincide with the topology at the time of the "attaching process". I guess this is what you asked for.

$\endgroup$
  • $\begingroup$ Thanks for the answer. 'it will always coincide with the topology at the time of the "attaching process"'. Ok, I agree. But and when we look to the topology induced by $X$? Taking $X_n$ for example, is it always true that the induced topology in $X_n$ by the limit topology (what I've called weak) is the original topology of $X_n$? Or this is true because we are attaching n-cells? $\endgroup$ – jngl123 Jun 14 '14 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.