Is path-connected a homotopy property of toplogical spaces? $X$ 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
 A: 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 $y_1$.
A: 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. 
