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Dec 8, 2014 at 16:08 comment added Andrej Bauer @MarcHoyois: that's completely irrelevant. Of course $A$ need not be a subspace of $X$, there's no way to make something into a subspace if it isn't (this is a case of set theory doing harm). The answer I give is the best possible: there is a canonical subspace $A'$ which is homeomorphic to $A$. Moreover, the pair of arrows $(r,f)$ is isomorphic to the pair of arrows $(f \circ r, \mathsf{incl}_{\mathsf{im}(f)})$ in the relevant category (whose objects are pairs of arrows $A \to X$ and $X \to A$ and morphisms are the evident ones).
Dec 8, 2014 at 14:14 comment added Mark Grant @MarcHoyois: I think Andrej's reading of the OP's definition of topological retract is "any space which is homeomorphic to a retract", and it seems that is what the OP intended. The fact that $A$ is given as a subset of $X$ is something of a red herring.
Dec 8, 2014 at 13:38 comment added Marc Hoyois This only shows that $A'$ is a topological retract of $X$, but $A$ need not be. See Mark Grant's counterexample!
Dec 8, 2014 at 13:17 vote accept Dominic van der Zypen
Dec 8, 2014 at 10:20 history answered Andrej Bauer CC BY-SA 3.0