# any cofibration $i:A \to B$ is a homeomorphism onto its image (question regarding the inverse map)

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:

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?

• $H_1^{-1}$ 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 $H_1(a)=x$. – Tyrone May 18 '20 at 16:24
• Hello Tyrone, yes exactly, that was my thought, too. I am a bit worried. – Zest May 18 '20 at 16:36
• $B$ 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. – Tyrone May 18 '20 at 16:57
• 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 $X$. – Zest May 18 '20 at 17:05
• 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. – 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.

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.

• 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! – Zest May 20 '20 at 17:42