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$X$ 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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $y_1$.

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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.

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  • $\begingroup$ 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

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