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?

  • $\begingroup$ $H_1^{-1}$ 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
  • 1
    $\begingroup$ 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

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.


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.

  • $\begingroup$ 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

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.