# Union of a sequence of increasing CW-complexes and Retract

I have updated my question based on the comments. I think I can solve the problem. But I hope you can help me verify it.

I am reviewing Algebraic Topology to prepare for my Qual. I have this problem

"Let $$X$$ be a cell complex and $$X_0\subset X_1\subset X_2\subset \cdots$$ a sequence of subcomplexes such that $$X=\cup_i X_i$$. Suppose that each $$X_i$$ is a retract of $$X_{i+1}$$. Prove that $$X_0$$ is a retract of $$X$$."

The materials for my past course is the book of Hatcher. I learned some materials about CW complexes like Whitehead's theorems. However, I don't think there are theorems about the relation between CW-complex and retracts.

For example, this question $X$ a CW complex is contractible if it's the union of an increasing sequence is clearly related to Whitehead's theorem. So unlike that, I do not know how to approach my problem.

Say $$f_i:X_i\to X_{i-1}$$ is the retraction. Then I construct $$f:\cup X_i\to X_0$$ where if $$x\in X_n$$, then we define $$f(x):=f_1 ...f_{n-1}f_n(x)$$. This map is well-defined. But how can I prove that it is continuous? So since $$X$$ is CW-complex, by weak topology, it suffices to show that $$f^{-1}(A)\cap X_i$$ is open in $$X_i$$. However, this is obvious since $$f^{-1}(A)\cap X_i=(f_1...f_i)^{-1}(A)$$ is an open set since $$f_1...f_i$$ is continuous.

I was wrong, I need $$f^{-1}(A)\cap S_i$$ is open in $$S_i$$ is the $$i$$-skeleton of $$X$$. Clearly I abused notations. So how can I go from $$X_i$$ to $$S_i$$?

• Oh okay, the weak topology. So it is obvious? I mean $f^{-1}(A) \cap X_i$ is just $(f_1...f_i)^{-1}(A)$, right? – Marcos G Neil Feb 5 at 6:49
• A subcomplex is by definition a union of some cells. And since $X=\bigcup X_i$ then $X$ has a weak topology relative to $\{X_i\}$ as well. – freakish Feb 5 at 11:57
• Can you quote that result? I do not know how to prove it – Marcos G Neil Feb 5 at 15:30
• A proof of Freakish's comment is basically contained in Hatcher's Proposition A.2 (around pg. 521 in the newest online edition). – Tyrone Feb 5 at 18:37

The retractions $$f_i : X_i \to X_{i-1}$$ give us retractions $$F_i = f_1 \circ \ldots \circ f_i : X_i \to X_0$$. Obviously $$F_{i+1} \mid_{X_i} = F_i$$. Thus we get a well-defined function $$F : X \to X_0$$ satisyfying $$F \mid_{X_i} = F_i$$. A function defined on a cell-complex is continuous iff its restriction to all closed cells is continuous. But each each closed cell $$c$$ is contained in some $$X_i$$, thus $$F \mid_c$$ is continuous.