*Question:* If $M$ is a compact smooth finite-dimensional manifold with boundary, is the inclusion of a closed subspace $A \subseteq M$ a cofibration? (I'm specifically interested in the case when $A$ is a smooth submanifold with boundary).


Does the following sketch proof work? *Sketch:* $M$ is homotopy equivalent to its interior which is a manifold with empty boundary, and since manifolds (I guess without boundary?) are absolute neighbourhood retracts (ANR's) and ANR's have the property that the inclusion of closed subspaces are cofibrations, the inclusion of $A \to M$ will be a cofibration.


I've seen similar statements for manifolds without boundary, for example when $A$ is a submanifold and both $A$ and $M$ have empty boundary, then it follows via Morse theory that $(X, A)$ have the homotopy type of a CW-pair and so the inclusion turns out to be a cofibration.


There is a potential answer to my question given on this site (<https://mathoverflow.net/a/16636/83360>), though I'm not sure if this applies to manifolds with boundary. Further, I haven't seen anything in the literature which covers the case when we consider manifolds with nonempty boundary (which is the reason for asking this question). If this has been shown somewhere in the literature, a reference would be very helpful.



If $M$ is a compact smooth finite-dimensional manifold with boundary, is the inclusion of a closed subspace $A \subseteq M$ a cofibration?
If the closed subset is locally compact and locally contractible then yes, the inclusion is a cofibration. This is surprisingly not very well known, but it follows from the classification of finite dimensional ANR's and the following fact from the answer of Tyrone [here](https://mathoverflow.net/questions/374160/is-the-alexander-horned-sphere-a-cofibration):



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> If $X$ is an ANR and $j:A\subseteq X$ is a closed subspace, then $A$ is an ANR if and only if the inclusion $j$ is a cofibration.
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