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If I have a function $p:\tilde X\to X$ and a function $f:Y\to X$ , then a function $\tilde f:Y\to\tilde X$ such that $p\circ\tilde f=f$ is called a lift of $f$ with respect to $p$. So a lift is just a name for a solution of the equation $p\circ x=f.$ Is there a name for the solution of the equation $x\circ p=f$ (with the functions' domains and codomains set so that it makes sense)?

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Maybe you are interested in "extension" problems, as opposed to lifting problems. – Fosco Loregian Mar 29 '13 at 13:08
up vote 4 down vote accepted

Its an extension.

The dual (with some relabelling) is: given a function $i : X \to \tilde X$ and a function $f : X \to Y$ we get a function $\tilde f : \tilde X \to Y$ such that $\tilde f \circ i = f$.

If $X \subseteq \tilde X$ and of $i$ is the inclusion map $X \hookrightarrow \tilde X$, this is just saying that $\tilde f$ is an extension of $f$ to $\tilde X$.

This notion of extension also makes sense when $i$ is an arbitrary injective function. It makes less sense when $i$ is not injective, but no less sense than the word lift when $p$ is not surjective.

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Why do I need $p$ to be surjective? I don't see anything like this here. – Bartek Mar 29 '13 at 13:12
You don't need $p$ to be surjective; if $p$ is not surjective, there are obvious functions which don't have lifts. – Thomas Andrews Mar 29 '13 at 13:15
Right, I've removed surjective/injective from my answer; this discussion of whether you need surjectivity for 'lift' to really mean 'lift' is distracting attention from the point that the most sensical dual to 'lift' is 'extension'. – Clive Newstead Mar 29 '13 at 13:19
An example is $\tilde X=CX$, the cone on $X$. Then $f:X\to Y$ is extensible to $CX$ if and only if $f$ is homotopy-equivalent to a constant map. – Thomas Andrews Mar 29 '13 at 13:26

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