Continuity of a function defined by cases on a space obtained as an attachment of n-cell along another space

Question

Reading "Morse Theory" by J.Milnor I found more than once this way of approach:

Let X and Y be two topological space. Suppose we want to define a continuous map from a space obtained by attaching a n-cell to X along a map $\phi: S^{n-1}\rightarrow X$. Let's call $X\sqcup_{\phi} D^{n}$ this space, $i:X\rightarrow X\sqcup D^{n}$ and $j:D^{n}\rightarrow X\sqcup D^{n}$ the two inclusion maps and $\pi:X\sqcup D^{n}\rightarrow X\sqcup_{\phi} D^{n}$ the natural projection. Additionally, suppose we know two continuous functions $f:X\rightarrow Y$ and $g:D^{n}\rightarrow Y$ such that the function $F:X\sqcup_{\phi} D^{n}\rightarrow Y$ defined by cases:

$F(\pi(i(x)))=f(x)$ if $x\in X$.

$F(\pi(j(x))=g(x)$ if $x\in D^{n}$

is well defined.

My problem concerns the proof of the continuity of the function: The book doesn't say anything about how to prove the continuity of the function defined this way. I tried to give an answer by applying pasting lemma: if I show that both the sets in which we defined F are closed, it follow the desired result. $\pi^{-1}\{\pi(i(x)): x\in X\}=\{(i(x)): x\in X\}\cup S^{n-1}$ is obviously a closed subspace of $X\sqcup D^{n}$, while I'm not able to say much about $\pi^{-1}\{\pi(j(x)):x\in D^{n}\}=j(D^{n})\cup i(\phi(S^{n-1}))$. If X is Hausdorff $\phi(S^{n-1})$ is closed in X and so $j(D^{n})\cup i(\phi(S^{n-1}))$ is closed in $X\sqcup D^{n}$ but in general I don't know how to prove it and in Milnor's book there aren't any hypothesis on X.

Can someone give me an hand? :)

Thanks in advance!

Answer 1

You're basically asking for the universal property of quotient spaces: If $f\colon A\sqcup B\rightarrow C$ is a continuous map with $f(a)=f(b)$ if $a\sim b$, then there is a continuous map $\tilde f\colon A\sqcup B\,/\!\sim\,\rightarrow C$.

To prove this, let $U\subseteq C$ be an open subset. By definition of the quotient topology, $\tilde f^{-1}(U)$ is open in $A\sqcup B\,/\!\sim$ if and only if $\bigcup \tilde f^{-1}(U)=(f|_A)^{-1}(U)\sqcup (f|_B)^{-1}(U)$ is open in $A\sqcup B$, which it is because $f$ is continuous.

In your case, $\bigcup \tilde f^{-1}(U) = \pi^{-1}(F^{-1}(U))=i(f^{-1}(U))\sqcup j(g^{-1}(U))$ is open because $i$ and $j$ are homeomorphisms when restricting the codomain to their corresponding image. In particular, they are open maps.

See also “Proof of the universal property of the quotient topology” and “Adjunction space is a pushout”.