When will "Retract $\iff$ Deformation Retract" hold true?

Problem. Given topological space $$X$$ and subspace $$A\subset X,$$ under what conditions will we be able to make the following claim: $$A$$ is a retract of $$X$$ if and only if $$A$$ is a deformation retract of $$X?$$

I know (from Hatcher's Chapter 0 exercises) there exist instances where rentract does not imply deformation retract. Seeing how abnormal these scenarios may be, though, I'm curious as to what extent are we able to say that deduce "contract if and only if deformation contract"? (Are there any theorems or statements anybody knows of immediately? That maybe I could use?) I've managed to come up with two statements, though I'm unsure if this is true:

(Edit: Immediately disproven. See comments.) Statement One. If $$X/(X-A)$$ is path-connected, then $$A$$ is a retract if and only if $$A$$ is a deformation retract.

Motivation. If we consider two objects in Euclidean Space, filled-in square $$X$$ with side lengths $$1$$ & disc with $$A\subset X$$ with radius $$1,$$ then it's immediately clear that there exists a homotopy (namely the linear homotopy) from the identity map $$\mathbb{1}_X$$ (on $$X)$$ to some retract $$r$$ from $$X$$ onto $$A.$$ The precise map for this homotopy could be defined via piecewise construction, where elements in $$A$$ are fixed and elements in $$X-A$$ follow there expected linear paths. If we were to entertain the scenario where the objects are less obviously "retract & deformation retract" (e.g. retract of a non-convex space), by property of path-connectedness, I believe would could still produce a homotopy which satisfies the desired conditions...however, this would need some proving. (And before I write up a proof, I would like to make sure there aren't any obvious counter-examples I'm missing.)

Statement Two. If $$X$$ is contractible, then $$A$$ is a retract if and only if $$A$$ is a deformation retract.

Motivation. This is a weaker variant of Statement One, and I propose it in the event that the previous statement isn't true. The motivation is shared from Statement One. For additional variants, one could consider the instance where $$X$$ is path-connected or simply connected (this would be more powerful than the contractibility condition).

What do you think? I appreciate any and all help I can get. As far as my understandning of Algebraic Topology is concerned, I have read through chatpers 0-2 of Hatcher, though I'm currently reviewing so as to refresh my memory and strength understanding.

• I don't think statement 1 could possibly be right, since it would hold for all path connected spaces $X$. Commented Jul 20 at 22:52
• @Randall You having restated my claim made me realize that my statement is very likely incorrect: Otherwise, the retracts will always have fundamental group isomorphic to the parent space (which I suspect a very easy counter-example could be constructed). I guess that just leaves the second statement then? Commented Jul 20 at 22:55
• @JAG131 Statement 1 is false: Take $X = S^n$ ($n > 0$) and $A$ to be any point in $X$. Then $A$ is a retract of $X$, but not a deformation retract. But the quotient $X / (X - A)$ is the Sierpinski space which is path connected. I don't see how the statement implies that retracts always have isomorphic fundamental groups, however. Commented Jul 20 at 22:57
• Also, the arguments you give aren't really arguments. I'd think about this a bit more before asking. Commented Jul 20 at 22:59
• @JAG131 A space for which all retracts are deformation retracts is contractible, yes. Any more interesting result will have to involve some conditions on $A$. Commented Jul 20 at 23:50

Statement 1 is almost never true. For example if $$A$$ is a point it is always a retract of $$X$$ (via the unique map $$X \to A$$) but if it's a deformation retract then $$X$$ must be contractible. Being a retract is a much weaker condition than being a deformation retract.