Recognizing an adjunction space

Question

I am currently studying Adjunction spaces using Brown's Topology and Groupoids. I am having trouble understanding exercise 4.5.3:

Let $B$ be a closed subspace of $Q$. For each $λ=1,\dots,n$, let $f_λ:X_λ→Q$ be a map, and let $A_λ$ be a closed subspace of $X_λ$ such that

1. $f_λ[A_λ]⊆B$,
2. $f_λ|X_λ\setminus A_λ$ is injective,
3. the sets $f_λ[X_λ\setminus A_λ]$ are disjoint and cover $Q\setminus B$,
4. $f_λ|X_λ$, $f_λ[X_λ]$ is an identification map.

Prove that a function $g:Q→Y$ is continuous if and only if $g|B$ ,$gf_λ$ is continuous, $λ=1,\dots,n$. Prove also that there is a homeomorphism $Q→B_{f1}⊔X1\dots_{fn}⊔Xn$ which is the identity on $B$.

I tried to prove that $Q$ has the final topology with respect to the inclusion $i_B:B\rightarrow Q$ and the maps $f_\lambda$. For that I take a subset $C$ in $Q$ whose inverse image for all these maps are closed, and I try to show that $C$ is closed. However, I am not able to prove it without adding the assumption that $f_\lambda[X_\lambda]$ is closed in $Q$ for each $\lambda$, and I have the feeling that it is necessary.

For example, take $Q=[0,2]$ and the closed subspace $B=[0,1]$ in $Q$. Take $X=(0,1)\cup (1,2]$ and the closed subspace $A=(0,1)$ in $X$. Let $f:X\rightarrow Q$ be the identity map. Then these spaces and maps satisfy the conditions above (do they, actually?). But the subset $C=(1,2]$ of $Q$ is not closed in $Q$ although $i_B^{-1}[C]$ is closed in $B$ and $f^{-1}[C]$ is closed in $X$. I cannot find where is the mistake in this example.

How can I solve this exercise without the additional assumption? Is my example wrong?

Answer 1

The space $Q$ has been defined in Exercise 4.5.1, thus your example does not really fit. But this is irrelevant as you could easily adjust your example.

In fact you gave a correct counterexample. Therefore the assertion is wrong as stated in the exercise.

By the way, you can construct a universal counterexample for $n=1$ as follows.

Let $Q$ be any space and $B$ a closed subspace which is not open. Then let $X = Q \setminus B$ and $A = \emptyset$. Consider the inclusion $i: X \to Q$. It clearly has properties 1. - 4. Now define $g : Q \to \mathbb R, g(q) = 0$ for $q \in B$ and $g(q) = 1$ else. This is not continuous because $B$ is not open. On the other hand $g\mid_B$ and $g \circ i$ are continuous.

Therefore additional requirements are needed. That $f_\lambda[X_\lambda]$ is closed in $Q$ clearly suffices. However, the weaker assumption that $\overline{f_λ[X_λ\setminus A_λ]} \subset f_\lambda[X_\lambda]$ will also do.