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Here is the problem: Let $X \subset Y$ be topological spaces with $X$ a deformation retract of $Y$. Then $Y$ is path connected if $X$ is path connected.

This seems simple, but I'm having trouble working out a proof. Here is what I'm thinking though: if $y,y' \in Y$ are two points, then under the deformation retract $r:Y \to X$, we can form a path $\varphi$ between $r(y) = x$ and $r(y') = x'$. The problem I run into is using the homotopy $f:Y \times I \to Y$ between $r$ and $\mathrm{id}_Y$, for I can't seem to "lift" the path $\varphi$ to $Y$.

Another thing I was thinking was using the isomorphism $i_*:\pi_1(X,x) \to \pi_1(Y,x)$ for any $x \in X$, and constructing a path from any point $y \in Y$ to a point $x \in X$ via this isomorphism, but I'm having trouble introducing a path to begin with.

This was a problem on an old topology qual. Hints are welcome. Thanks!

Ben Tighe
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4 Answers4

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Let $A$ be a deformation retract of $X$. Then, let $a,b \in A$. Clearly there is a path $\lambda:[0,1] \to X$ that connects $A,B$. However, since there is a homotopy between $id:X \to X$ and $i: A \hookrightarrow X$, we can take $F:X \times [0,1] \to A$ with $F \mid_A$ being identity.

Take the inclusion $\lambda \times id :A \times [0,1] \hookrightarrow X \times [0,1],$ and consider $F \circ (\lambda \times id)$. Then $F_1$ (basically at $X \times \{1\}$ has $A$ as its image, was continunous, and since it fixed elements of $A$, we obtain a path from $a$ to $b$ contained in $A$.

Alternatively, as Ted Shifrin mentions in the comments, $i_*:H_0 \to H_0$ is an isomorphism, so path components of $A$ and $X$ should be the same.

Andres Mejia
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  • Sorry my notation was so cumbersome, I hope that the idea is clear. – Andres Mejia Dec 08 '17 at 02:38
  • No, I get the point. My trouble was constructing the path, and this seems to do the trick. Thanks! – Ben Tighe Dec 08 '17 at 02:43
  • You have unnecessarily made things complicated by introducing cumbersome notations. This is quite straightforward. It can be answered in fewer lines. – Akiro Kurosawa Aug 22 '23 at 20:57
  • Given the level of the question I actually think that this answer is the right level of concreteness. If you believe that there is a better way to express the solution feel free to write an answer. – Andres Mejia Aug 24 '23 at 00:32
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More generally one can show that the assignment $\pi_0 : hTop_* \to Set$ is a functor, where $\pi_0(X,x)$ is the set of homotopy classes of maps $(\mathbb{S}^0,0) \to (X,x)$.

To show $\pi_0$ is well-defined, one first needs to know that if $f,g: X \to Y$ are homotopic, then $\pi_0(f)=\pi_0(g)$. Let $F: X \times I \to Y$ be a homotopy between $f$ and $g$, and let $\phi: \mathbb{S}^0 \to X$ be continuous. Define $G: \mathbb{S}^0 \times I \to Y$ by $G(s,i)=F(\phi(s),i)$. Then $G(s,0)=F(\phi(s),0)=(f \circ \phi)(s)$ and $G(s,1)=F(\phi(s),1)=(g \circ \phi)(s)$. Hence $G$ is a homotopy and so $f$ and $g$ represent the same map $[\mathbb{S}^0, X] \to [\mathbb{S}^0, Y]$, i.e. $\pi_0(f)=\pi_0(g)$.

Once we have this, it's easy to deduce that $\pi_0(f \circ g)= \pi_0(f) \circ \pi_0(g)$ for any maps $g: X \to Y$ and $f: Y \to Z$, and thus that, for example, the path components of a space and those of its deformation retract are in bijection.

Note the same proof works to show that for any space $X$, the assignment $[X,-]: hTop_* \to Set$ is a functor (in particular that the $\pi_n$'s are all functors).

Exit path
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  • In leibnewtz' proof the functor $\pi_0$ is defined on the category $hTop_\ast$ of pointed spaces and pointed homotopy classes of basepoint-preserving maps. It shows that a pointed homotopy equivalence $f$ induces an isomorphism $\pi_0(f)$. However, in general (free) homotopy equivalences are no pointed homotopy equivalences so that there is a little gap in the proof (which can easily be filled). It is more elementary to work with the functor $\pi : hTop \to Set$ given by $\pi/X) =$ set of homotopy classes $\ast \to X$ where $\ast$ is a one-point space. – Paul Frost Jun 17 '18 at 09:23
  • @PaulFrost You raise a good point. But in the case of interest, a retraction $Y \to X$ can be thought of as a pointed homotopy equivalence by fixing a suitable base point in $X$. – Exit path Jun 17 '18 at 15:18
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    @eibnewtz If you have a retraction $r : Y \to X$, then $\pi_0(Y,x)$ dominates $\pi_0(X,x)$ which shows that $X$ is path connected if $Y$ is. For the converse we need something more more. If $r$ is a strong deformation retraction, it is a pointed homotopy equivalence and we are done. If it is only a deformation retraction ($i \circ r \simeq 1_Y$ not necessarily rel. $X$), we cannot be sure about this. However, working with $\pi(X)$ as above will do. Clearly there is a bijection $\pi(X) \approx \pi_0(X,x)$ for any choice of $x$ (homotopy equivalences induce bijections between path components). – Paul Frost Jun 17 '18 at 17:17
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The accepted answer seems to show the opposite implication: namely, $Y$ path connected implies $X$ path-connected. However, it is easy to see there is an equivalence.

In what follows, let $r:Y\rightarrow X$ be the retraction. First, the easy direction:

If $Y$ is path-connected, then $X$ is path-connected.

Proof: $X=r(Y)$ is the image of a path-connected set. More to the point, if $p:[0,1]\rightarrow Y$ is a path from $x_1$ to $x_2$, then $r\circ p$ is a path in $X$.

If $X$ is path-connected, then $Y$ is path-connected.

Proof: Let $y_1, y_2\in Y$. Also let $H:Y\times[0,1]\rightarrow Y$ be the deformation retract. Then the "vertical slice" $H(y_1,-):[0,1]\rightarrow Y$ is a path from $H(y_1,0)=y_1$ to $H(y_1,1)=r(y_1)$. Similarly, there's a path from $y_2$ to $r(y_2)$. This we have a concatenation of paths $$ y_1\rightarrow r(y_1)\rightarrow r(y_2)\rightarrow y_2 $$ where the middle path exists because $X$ is path-connected. Thus there's a path from $y_1$ to $y_2$.

Steve D
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Let $H:X\times I\rightarrow X$ by the homotopy for which $H(x,0)=x$, $H(x,1)\in A$ for each $x\in X$ and $H(a,t)=a$ for each $a\in A$ and each $t\in I$.

Now, let $a_1,a_2\in A$. Since $X$ is path connected, there exists a path $\gamma:I\rightarrow X$ for which $\gamma(0)=a_1$ and $\gamma(1)=a_2$. Now define $r:X\rightarrow A$ by $r(x)=H(x,1)$.

$r\circ \gamma$ is the desired path in $A$.

Moe Khaled Bin-Lateef
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  • What you have shown here is the converse of what OP asked i.e. if $X$ deformation retracts to $A$ then $A$ is path connected as long as $X$ is path connected. For answering OP's question we should first note that if $X$ deformation retracts to $A$ then there is a homotopy between $\text {id}_X$ and a retract from $X$ onto $A.$ This allows us to join $x$ and $r(x)$ by a path in $X.$ Now for two points $x$ and $y$ In $X$ we know that $r(x)$ and $r(y)$ are connected by a path in $A.$ So in order to get a path between $x$ and $y$ we concatenate the paths between $x$ and $r(x)$ in $X$...contd – Akiro Kurosawa Aug 22 '23 at 20:50
  • ...followed by a path between $r(x)$ and $r(y)$ in $A \subseteq X$ followed by a path between $r(y)$ and $y$ in $X.$ Pretty straightforward. Right? – Akiro Kurosawa Aug 22 '23 at 20:52