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If $f:A \rightarrow X$ is a continuous map between CW-complexes, then is $f$ necessarily a cofibration? I know that when $A$ is a subcomplex of $X$ and $f$ is the inclusion, the conclusion is true. Also, when $f$ is cellular, we can use the mapping cylinder to show the conclusion is true. But how about the general map $f$?

Stefan Hamcke
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Summer
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2 Answers2

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Why an answer after so many years? Because there are some more aspects.

The minimal requirement for $f : A \to X$ to be a cofibration is that it is a closed embedding (all cofibrations are embeddings, and if $X$ is Hausdorff then $f(A)$ is closed in $X$).

Now have a look at Is a closed embedding of CW-complexes a cofibration?.

You will see that it true at least if $X$ is locally finite.

Paul Frost
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A cofibration is a closed inclusion (injective with closed image) for Hausdorff spaces and so not all continuous maps between CW-complexes are cofibrations. Example $f\colon\{1,2\}\rightarrow\{1\}$.

Dan Rust
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