Welcome to today’s dose of algebraic topology! I’ve just started reading the amazing book by J.P.May. I don’t plan to post all exercises, but there’s many things so thrilling I can’t resist writing down~ qwq
Subtitle: Commutative Diagram Test Post
The category we work in is $\mathrm{CGWH}$, i.e., the full subcategory of $\mathrm{Top}$ with objects the weak Hausdorff k-spaces.
It has a structure of internal $\operatorname{Hom}$: We give $Y^X:=\operatorname{Hom}(X,Y)$ the k-ification (the right adjoint of forgetful $\mathrm{kTop}\hookrightarrow\mathrm{Top}$) of the compact-open topology (given by the subbase $ \{U^K:K\text{ compact}, U\text{ open}\}$). Then the category is cartesian closed with respect to it: For any $X,Y,Z\in\mathrm{CGWH}$, \(Z^{X\times Y}\simeq(Z^Y)^X.\)
Therefore a homotopy $X\times I\to Y$ is the same as a map $X\to Y^I$.
We define a cofibration to be a map $i:A\to X$ such that in the following diagram,
if the “square” $h\circ\iota_0=f\circ i$ commute, then there exists some $\tilde h$ that extends $h$. This is called the homotopy extension property.
Notice that we essentially only need one “testing” $X\xrightarrow{f}Y\xleftarrow{h}A\times I$ for this property, namely, the pushout $Mi:=X\underset{A}{\cup}(A\times I)$. If $\tilde h:X\times I\to Mi$ exists, we can just compose with it the “pushout” map $Mi\dashrightarrow Y$.
Now if we look at the diagram:
with some attention, we see that
\[\begin{equation} r\circ j=\operatorname{id}_{Mi} \end{equation}\]which triggers,
left as an exercise, it follows that a cofibration is an inclusion with closed image.
Well this is not at all obvious to me. I proved it in several steps. The obvious is
Proposition 1. The cofibration $i$ is injective.
Proof. $j$ has a retraction by (1), so it’s injective, and so is $\left.j\right|_{A\times\{1\}}$, which is just $i$.
Proving the image is closed caused a headache. Finding the following result slightly alleviates it.
Lemma. Any continuous map $j:A\to X$ with a retraction $r:X\to A$ is a topological embedding. So we just identify $j(A)$ with $A$.
Proof. It’s injective, and for $U\underset{\text{open}}{\subset}A$, clearly
\[j(U)=r^{-1}(U)\cap j(A).\]
Propostion 2. If $X$ is Hausdorff, then $j$ defined as above has closed image.
Otherwise, we have some $x\in\overline{A}\setminus A$. Since $r(x)\in A$, $x\neq r(x)$. Therefore we can find disjoint open neighborhoods $U$, $V$ separating them. Therefore,
\[A\cap U\cap r^{-1}(V)\neq\varnothing,\]which is ridiculous~
Proposition 2.’ If $X$ is CGWH, then $j$ defined as above has closed image.
By weak Hausdorff, we only need that given any compact hausdorff subspace $K$ of $X$, $A\cap K$ is closed. Again, suppose $x\in\overline{A\cap K}\setminus A\subset K$. Notice that, by continuity of $r$,
\[\begin{align*} r(\overline{A\cap K}\setminus A) &\subset r(\overline{A\cap K}) \\ & \subset \overline{r(A\cap K)} \\ &=\overline{A\cap K} \\ &\subset K. \end{align*}\]So $x$, $r(x)\in K$, and we can proceed as before. Take $U\cap K$, $V\cap K$ separating them, i.e., $U\cap K\cap V=\varnothing$. However,
\[\left(A\cap K\right)\cap\left(U\cap r^{-1}(V)\right)\neq\varnothing.\]
Q.E.D. (Quite Expensive Decoration)
Post date: 2024/11/05
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