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A deformation retraction of $X$ onto a subspace $A$ is a homotopy $H:X\times [0,1]\to X$ such that $H(x,0)=x$, $H(x,1)=r(x)\in A$, and $H(a,t)=a$ for every $a\in A$ and every $t$. Thus the retract is fixed pointwise throughout. With inclusion $i:A\hookrightarrow X$, the equations $r\circ i={\mathrm{id}}_A$ and $i\circ r\simeq {\mathrm{id}}_X$ exhibit a homotopy equivalence. See Hatcher, Algebraic Topology, Chapter 0, pp. 2–3.

Take the closed annulusAn exact deformation retraction onto its inner circle isThe denominator is nonzero everywhere on the annulus. If the starting radius is $\rho =\Vert x\Vert$, its radius at time $t$ is $(1-t)\rho +t$. This lies between $1$ and $\rho \le 2$, verifying that the entire homotopy stays in the original annulus. When $\rho =1$, the point is stationary, as required.

For $x=(0,2)$, times $0,1/2,1$ give $(0,2),(0,3/2),(0,1)$. A radius-$3/2$ loop similarly becomes a radius-$5/4$ loop halfway through, then the unit circle. Its angle and winding number are preserved. This is useful because computing the fundamental group of the annulus reduces to computing that of the circle; the explicit homotopy equivalence establishes why the answer transfers.

Use the mapping cylinder convention $(x,1)\sim f(x)$ for $f:X\to Y$. Its bottom copy of $X$ is a retract precisely when $f$ has a left homotopy inverse $g:Y\to X$. A retraction restricted to $Y$ supplies $g$, and its values along the cylinder supply the homotopy from ${\mathrm{id}}_X$ to $gf$. Conversely use that homotopy on the cylinder and $g$ on $Y$; agreement at the attached end makes the map descend by the universal property of the quotient topology.

Choose a small skeleton $\mathcal{F}$ of the finite spectra, including every integer suspension. There is such a set of representatives: a finite list of stable cells and their attaching maps specifies a finite cell model, and the possible attaching maps form sets. For every $F\in \mathcal{F}$ and every stable map $v:F\to D$, take one copy of $F$ and formDefine the evaluation map $e:X\to D$ by taking its restriction to the $(F,v)$ summand to be $v$. Complete it to the cofiber sequence of spectraThis is the universal evaluation phantom map construction.

First, $p$ is nonzero. If $p=0$, applying $[D,-{]}_{\mathrm{st}}$ to the cofiber sequence of spectra gives the exact portionSince $p_{\ast }(1_D)=p=0$, exactness supplies $s:D\to X$ with $es=1_D$. Thus $D$ is a retract of $X$. This is also a wedge summand: complete $s$ to a cofiber sequence of spectra $D\to X\to C^{\prime }\to \Sigma D$. The left inverse $e$ makes this triangle split, giving $X\cong D\vee C^{\prime }$ in the stable homotopy category. One can see the splitting directly from exactness: the connecting map is zero because $e$ extends $1_D$ across $s$, and consequently the quotient map $X\to C^{\prime }$ admits a section. This contradicts the assumption on $D$. Hence $p\ne 0$.

For any finite spectrum $F$ and any $v:F\to D$, its chosen representative has a summand inclusion ${\iota }_v:F\to X$ with $e{\iota }_v=v$, after transferring along the representative isomorphism. Consecutive maps of a cofiber sequence of spectra compose to zero, soThus $p$ is a phantom map of spectra. To establish the requested conclusion on all spaces, rather than just coefficient groups, let $K$ first be a finite based CW complex. Put $A={\Sigma }^{\infty }K$. The Spanier-Whitehead dual $A^{\vee }=F(A,\mathbb{S})$ is a finite spectrum. The evaluation and coevaluation adjunction of Spanier-Whitehead duality identifiesUnder this identification $p_{\ast }$ is postcomposition with $p$. Its source is a finite spectrum, including when $k$ is negative, so the calculation $pv=0$ proves that $p_{\ast }$ is zero on ${\tilde{D}}_k(K)$ for every $k$.

Now let $Y$ be an arbitrary based CW complex. It is the filtered union, or filtered homotopy colimit of spaces, of its finite based subcomplexes $K$. The suspension spectrum functor and smash product of spectra preserve that homotopy colimit. The shifted sphere spectrum is compact, so taking stable homotopy groups givesEvery element of the first group therefore comes from a finite $K$, where $p_{\ast }$ has just been proved zero. Consequently $p_{\ast }$ is zero on $Y$ as well. A based CW complex replacement extends the conclusion to arbitrary based spaces in the homotopy category. We have constructedThis uses continuity of represented homology theory, not a continuity assertion for represented cohomology theory; the latter need not hold.

A subspace $Y\subseteq X$ is a retract of $X$ if there is a continuous retractionsuch that $r(y)=y$ for every $y\in Y$. A space is a contractible space if its identity map is homotopic to a constant map.

Let $i:Y\hookrightarrow X$ be the inclusion and letbe a contraction, with $H(x,0)=x$ and $H(x,1)=x_0$. Thenis continuous and satisfiesThus $K$ contracts $Y$, proving that every retract of a contractible space is contractible.

Yes. Take $X=S^1$, the circle, and $A=\{1\}\subset S^1$. The constant function $r(z)=1$ is a continuous map and satisfies $ri={\mathrm{id}}_A$, so $A$ is a retract of $X$.

However, their fundamental groups areA homotopy equivalence between path-connected spaces induces a group isomorphism of their fundamental groups, allowing the usual change of base point. The infinite cyclic group is not the trivial group. Therefore a retract need not have the homotopy type of its ambient space.

Put $A=T^2\setminus B^2$ and let $E=\partial A=S^1$ denote the attaching circle. The version of the Seifert-van Kampen theorem used here states that if $X=U\cup V$ is an open cover with $U,V,U\cap V$ path-connected and a common base point in $U\cap V$, thenThe expression on the right means the pushout of groups of the two inclusion-induced group homomorphisms; they need not be injective.

Choose a thin collar neighbourhood $C_A$ of $E$ inside $A$, and a thin equatorial band $C_S$ inside $S^2$. For example, before the square's side identifications take $C_A=\{v:1\le |v|<1+ϵ\}$ with $0<ϵ<1/2$, and take $C_S=\{(x,y,z)\in S^2:|z|<ϵ\}$. In the glued topological space setThese are open subsets: their preimages in the two pieces are open and respect the gluing. They cover $X$, and the three deformation retractionsfix $E$. Thus all three spaces are path-connected, and a base point $b\in E$ is preserved.

For completeness, the fundamental group of a once-punctured torus is $F(a,b)$, and its boundary loop of a once-punctured torus represents $[a,b]=ab{a}^{-1}{b}^{-1}$, up to orientation and a change of base point. Here is the geometry behind both facts. In the square model, move each point of $A$ radially outwards to the square boundary:The denominator never vanishes on $A$. The homotopy stays outside the missing open disk, stays inside the square, and fixes the square boundary, so it descends through the side identifications. The resulting retract is a wedge of two circles. Its fundamental group is the free group on the two oriented side based loops $a,b$. The inner boundary loop travels around the square once under this deformation retraction; its successive sides read $a,b,a^{-1},b^{-1}$.

The sphere $S^2$ is simply connected. Hence the group homomorphism from ${\pi }_1(E)\cong \mathbb{Z}$ to ${\pi }_1(V)$ is trivial, while its map to ${\pi }_1(U)$ sends a group generator to $[a,b]$. Applying the Seifert-van Kampen theorem givesThe final group isomorphism follows because the single relator makes $a,b$ commute, leaving the free abelian group on two group generators. The two hemispheres kill the same boundary loop; they do not create another group generator.

Choose one representative of each finite spectrum and form $X={\bigvee }_{(F,u:F\to D)}F$. Let $e:X\to D$ evaluate each summand and let $p:D\to \mathrm{cofib}(e)$. Every map from a finite spectrum factors through its corresponding summand, so $p$ is a phantom map of spectra. If $p=0$, exactness supplies a section of $e$, making $D$ a retract of $X$. Consequently a spectrum that is not a retract of any wedge of finite spectra has a nonzero such phantom.