$X$ and $Y$ are homotopy equivalent so there are maps $\alpha: X \rightarrow Y$ and $\beta : Y \rightarrow X$ whose composites satisfy : $\beta\alpha \simeq id_X$ and $\alpha\beta \simeq id_Y$

$X$ is path connected if all points $a$ and $b$ be connected by paths $p:[0, 1] \rightarrow X$ such that $p(0)=a$ and $p(1)=b$.

Do we need to show that $Y$ is also connected?

Thanks for your help

  • $\begingroup$ Homotopy is not a relation between spaces but between maps : what are f and g here ? $\endgroup$ – Captain Lama May 19 '16 at 20:10
  • $\begingroup$ f and g are two maps which are homotopic to one another, $f, g : X \rightarrow Y$ $\endgroup$ – thinker May 19 '16 at 20:13
  • $\begingroup$ I think I have made a confusion with maps and spaces, I will edit my question now $\endgroup$ – thinker May 19 '16 at 20:14
  • 1
    $\begingroup$ I made an edit to correct some terminology: Spaces are called homotopy equivalent in the given case for $X$ and $Y$. $\endgroup$ – Alex G. May 19 '16 at 20:20

Something which is generally true is that a homotopy equivalence $\alpha: X\to Y$ induces isomorphisms on all homotopy groups. That is, $\alpha_*: \pi_n(X) \to \pi_n(Y)$ is an isomorphism for all $n$. In the case $n=0$, $\pi_0$ is just a set, not a group, and $\alpha_*$ is a bijection. Thus $X$ is path connected iff $\#\pi_0(X) = 1 \iff \#\pi_0(Y) = 1$ iff $Y$ is path connected.

I haven't told you why this is true though, and really, that is what I should do. Given $y_0, y_1 \in Y$, consider $\beta(y_0), \beta(y_1) \in X$, and let $\gamma:I\to X$ be a path from $\beta(y_0)$ to $\beta(y_1)$. Then $\alpha \circ \gamma:I \to Y$ is a path from $\alpha\circ\beta(y_0)$ to $\alpha\circ\beta(y_1)$. Additionally, the homotopy $H: Y\times I \to Y$ from $\alpha\circ \beta$ to $id_Y$ defines a path from $\alpha\circ\beta(y_0)$ to $y_0$, and likewise for $y_1$. Concretely, the path is $t\mapsto H(y_0, t)$. Thus, we may compose these paths to get one from $y_0$ to $y_1$.


Yes. The zeroth singular homology group $H_0(Y)$ is defined to be the free abelian group generated by all the elements of $Y$, modulo an equivalence relation in which two generators (points in $Y$) are equivalent if there is a continuous path connecting them. So in the quotient, we will have exactly one generator for each path component.

You are assuming that $X$ is path connected, so since any two points can be joined by a continuous path, there is exactly one free generator of $H_0(X)$ (and so $H_0(X) \simeq \mathbb{Z}$). Now a homotopy equivalence induces an isomorphism on homology groups, so we have that $H_0(Y) = \mathbb{Z}$. Unraveling the definition again we can see that this means that any two points in $Y$ can be joined by a continuous path.

  • $\begingroup$ To use this, one must work out why homology induces the same homomorphism on homotopic maps. Is this really easier than directly showing that a homotopy equivalence preserves then number of path components? $\endgroup$ – Alex G. May 19 '16 at 20:47
  • $\begingroup$ I agree, your answer is better since it just uses the basic definitions and is still very simple, this was just the first thing I thought of (I upvoted your answer for this reason). Submitted it before I saw you had answered the question. $\endgroup$ – Charlie Cifarelli May 19 '16 at 20:50
  • $\begingroup$ I think Alex's answer is correct on $\pi_0$ but it should be noted that homotopy groups $\pi_n, n>0$ are defined not for spaces but for spaces with base point, This distinction is sometimes elided in expositions, but is crucial. $\endgroup$ – Ronnie Brown May 19 '16 at 21:27
  • $\begingroup$ @RonnieBrown You are correct. I was purposefully ignoring that point to avoid getting off topic. $\endgroup$ – Alex G. May 20 '16 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.