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I was recently working on a problem that introduced the homotopy extension property as a cofibration $i:A \to B$. Let's say we are given the commutative diagram:

enter image description here

Now, if $i:A \to B$ is the inclusion map, the cofibration property is precisely the homotopy extension property. I was trying to understand the given solution as to why any cofibration is a homeomorphism onto its image.

The solution says:

Consider the mapping cylinder of $i$ which is $X = (A \times I)\sqcup B/\sim$, where $(a,0)$ gets glued to $i(a)$. We set $H$ to be the obvious projection $A \times I \to X$ and $G_0$ to be the inclusion $B \to X$. By the cofibration property there is a map $G$ as in the diagram, and denote by $G_1: B\to X$ the composition of $G$ with the inclusion of $B$ at height $1$. Define $H_1:A \to X$ analogously. By construction, $H_1$ is a homeomorphism onto its image, and $H_1 = G_1 \circ i$. Thus $H_1^{-1}$ is defined.

I have issues seeing why $H_1^{-1}$ is defined. Could someone please elaborate? Why and how is $H_1^{-1}$ defined? What is the key observation here?

Edit: i have added the last paragraph of the solution.

Thus $H_1^{-1}$ is defined and continuous on the image of$G_1|_{i(A)}$. It follows that $H_1^{-1}\circ G_1|_{i(A)}: i(A) \to A$ is a continuous inverse to $i$. Thus $i$ is in particular injective and also surjective (onto its image) so all the maps arehomeomorphisms.

As being mentioned in the comments, $H_1^{-1}$ is only defined on the image of $H_1$. Is the solution not correct or where am i being mistaken?

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  • $\begingroup$ $H_1^{-1}$ is only defined on the image $H_1(A)\subset X$. In particular $(H_1^{-1})(x)=a$, where $a\in A$ is the unique element with $H_1(a)=x$. $\endgroup$ – Tyrone May 18 '20 at 16:24
  • $\begingroup$ Hello Tyrone, yes exactly, that was my thought, too. I am a bit worried. $\endgroup$ – Zest May 18 '20 at 16:36
  • $\begingroup$ $B$ is embedded in $X$ as a closed subspace. Thus $X\setminus B$ is an open set homeomorphic to $A\times(0,1]$. This is clear from the quotient topology on $X$. In particular $H_1$ maps $A$ homeomorphically onto $A\times\{1\}\subset X$. The inclusion $in_1:A\hookrightarrow A\times(0,1]$ is an embedding since it has a retraction. Namely $pr_1:A\times(0,1]\rightarrow A$. Thus $H_1$ is an embedding. $\endgroup$ – Tyrone May 18 '20 at 16:57
  • $\begingroup$ Thank you, but having $H_1$ being an embedding just tells me that $H_1^{-1}$ is only defined on the image of $H_1$, doesnt it? In other words, i can't say $H_1^{-1}$ is defined on $X$. $\endgroup$ – Zest May 18 '20 at 17:05
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    $\begingroup$ I think i got it, thanks Tyrone. I think my issue was that i failed to chase the diagram properly and i got stuck at $H_1^{-1}$ is defined, not realizing, that the actual claim was: $H_1^{-1}$ is defined and continous on the image of $H_1$ which is the key observation. $\endgroup$ – Zest May 18 '20 at 17:29
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I think Tyrone's comments have clarified that "$H_1$ is a homeomorphism onto its image" means that $H'_1 : A \stackrel{H_1}{\to} H_1(A)$ is a homeomorphism which is obvious by definition of $H$ (note that $H_1(A)$ is the image of $A \times \{1\}$ under the quotient map ($A \times I) + B \to X$). Since certainly $H_1 = G_1 \circ i$, we get $H_1(A) = G_1(i(A))$ so that $G_1$ restricts to $G'_1 : i(A) \stackrel{G_1}{\to} H_1(A)$. By construction we have $G'_1 \circ i = H'_1$.

Let $i' : A \stackrel{i}{\to} i(A)$ (which is a continuous surjection) and $\phi = (H'_1)^{-1} \circ G'_1$. Then $$\phi \circ i' = (H'_1)^{-1} \circ G'_1 \circ i' = id_A .$$ This shows that $i'$ must be injective. Thus $i'$ is a continuous bijection and $\phi$ is its inverse which is continuous. Therefore $i'$ is a homeomorphism which means that $i$ is an embedding.

Note that an alternative proof can be based on $X = A \times I / A \times \{0\}$ which is a variant of the cone on $A$. Let $H : A \times I \to X$ be the quotient map and $G_0 : B \to X$ be the constant map $G_0(b) \equiv *$, where $*$ is the equivalence class of $A \times \{0\}$. Now argue as above.

Edited:

Let us try to understand why the mapping cylinder $X$ occurs in the above proof. What follows is perhaps a little more complicated, but I hope it makes it more transparent.

For a space $Z$ let $i^Z_t : Z \to Z \times I, i^Z_t(z) = (z,t)$. This is an embedding for each $t \in I$. The mapping cylinder $X = M(i) = \left((A \times I) + B \right)/(a,0) \sim i(a)$ is the pushout of the pair of maps $i^A_0 : A \to A \times I$ and $i : A \to B$. It comes along with maps $H : A \times I \to X$ and $G_0 : B \to X$ such that $G_0 \circ i = H \circ i^A_0$ which satisfy the universal property of the pushout. These maps are those occurring in your diagram. They are the restrictions of the quotient map $q : (A \times I) + B \to X$ to $A \times I$ and to $B$. Since $(i \times id_I) \circ i^A_0 = i^B_0 \circ i$, there exists a unique map $F : X \to B \times I$ such that $F \circ H = i \times id_I$ and $F \circ G_0 = i^B_0$. Explicitly it is given by $F([a,t]) = (i(a),t)$ and $F([b]) = b$.

Since $i$ is a cofibration, we moreover find a (non-unique) map $G : B\times I \to X$ as in your diagram. By the universal property of the pushout we have $G \circ F = id_X$ because $(G \circ F) \circ G_0 = G \circ i^B_0 = G_0 =id_X \circ G_0$ and $(G \circ F) \circ H = G \circ (i \times id_I) = H =id_X \circ H$. Thus $F$ is an embedding. In fact, each map $e : Y \to Z$ which has a left inverse $r : Z \to Y$ (which means $r \circ e = id_Y$) is an embedding: Clearly $e$ must be injective so that the map $e' : Y \stackrel{e}{\to} e(Y)$ is a continuous bijection with $(e')^{-1} = r\mid_{e(Y)}$ which is continuous.

The map $j_1 : A \stackrel{H \circ i^A_1}{\to} A' = H(A \times \{1\}) \subset X$ is a homeomorphism. We have $j_1(a) = [a,1]$. Trivially $k_1 : i(A) \to A'' = i(A) \times \{1\} \subset B \times I$ is a homeomorphism. Since $F$ is an embedding and $F(A') = F(H(A\times\{1\}) = (i \times id_I)(A\times\{1\}) = i(A) \times \{1\} = A''$, we see that $F' : A' \stackrel{F}{\to} A''$ is a homeomorphism. But for $i' : A \stackrel{i}{\to} i(A)$ we have $F' \circ j_1 = k_1 \circ i'$ which implies that $i'$ is a homeomorphism.

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  • $\begingroup$ Thanks for the insights, Paul. I will get back to this very soon and take a look at it, i might return with a question or two. Thanks! $\endgroup$ – Zest May 20 '20 at 17:42

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