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By a standard technique of inductive killing everything relevant (in this case decreasing homeomorphisms between uncountable $G_\delta$-subsets of the real line) it is possible to prove the following fact.

Theorem (CH). Under CH the real line contains an uncountable subset $X$ admitting no strictly decreasing function $f:Z\to X$, defined on some uncountable subset $Z$ of $X$.

On the other hand, a known PFA-results of Baumgartner (about the order isomorphness of any $\aleph_1$-dense subsets of the real line) implies the following

Theorem (PFA). Under PFA, for any uncountable subset $X\subset\mathbb R$ there exists a strictly decreasing function $f:Z\to X$, defined on some uncountable subset $Z\subset X$.

Now

Question. Can this PFA-theorem be proved under a weaker assumption like OCA or (MA$+\neg$ CH)?

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    $\begingroup$ "defined on an": do you mean "defined on any"? the article "a" is ambiguous, especially in such sentences. $\endgroup$
    – YCor
    Commented Jan 22, 2019 at 18:38
  • $\begingroup$ @YCor Thank you for the comment. I changed "an" to "some". $\endgroup$
    – Taras Banakh
    Commented Jan 22, 2019 at 18:45
  • $\begingroup$ OK now the meaning is clear. Anyway the correct English is "any", since the sentence is in the negative. $\endgroup$
    – YCor
    Commented Jan 22, 2019 at 18:58

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In Todorcevic's book "Partition Problems in Topology" I have found Proposition 8.4(c) saying that under OCA for any uncountable sets $X,Y$ of reals there exists a strictly increasing function $f:Z\to Y$ defined on some uncountable subset $Z$ of $X$.

This proposition implies that for any uncountable set $X\subset \mathbb R$ there exists a strictly increasing function $f:Z\to -X$ defined on some uncountable subset $Z\subset X$. Then the function $-f:Z\to X$ is stricly decreasing.

Therefore the PFA-theorem in OP holds under OCA.

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    $\begingroup$ To complement this answer, I think the answer should be no for $\mathsf{MA}$. Abraham and Shelah proved that $\mathsf{MA}$ doesn't give you Baumgartner's theorem by showing $\mathsf{MA}$ is consistent with the existence of something called an entangled set of reals. I think an entangled set shouldn't admit the sort of map you're describing. (I don't have time to check the details today, which is why I'm just writing a comment -- I hope this helps.) $\endgroup$
    – Will Brian
    Commented Jan 29, 2019 at 14:10
  • $\begingroup$ @WillBrian Thank you for the comment. I will try to check the paper of Abraham-Shelah. $\endgroup$
    – Taras Banakh
    Commented Jan 29, 2019 at 15:03

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