If f~ g then $ \pi_0(f) =\pi_0(g)$

Question

This question was left as an exercise in my class of Algebraic Topology and I am struck on on of it parts.

So, I am posting it here.

Question: For all topological spaces X, we let $\pi_0(X) $ denote the set of arcwise connected components of X. We let $\bar{x}$ denote the arcwise component of $x\in X$. For all continuous maps $f : X\to Y$ , then $\pi_0(f) : \pi_0(X) \to \pi_0(Y)$ is given by $\overline{x}\to \overline{f(x)}$.

Then show that (1)(a) $\pi_0(id_X)=id_{\pi_{0}(X)}$.

(b) $\pi_0( g \circ f)= \pi_0(g)\circ \pi_0(f)$

(2) If f~g , then $\pi_0(f)= \pi_0(g)$.

I have done 1 but with 2 I am not sure how exactly I should use the condition that f~g. f~g implies that there exists $H: X\times I\to Y$ such that $H(x,0)=f(x) $ and $H(x,1)=g(x) $ for all $x\in X$, here $H$ is continuous. $\pi_0(f)=\overline{f(x)}$ and $\pi_0(g)= \overline{g(x)}$.

But I am not sure how can I prove them to be equal. Can you please help me with it?

Answer 1

I'm going to assume that arcwise-connected here means the same as path-connected.

We want to show that if $f \sim g$, then $\pi_{0}(f) = \pi_{0}(g)$. By definition, we have $\pi_{0}(f)(\overline{x}) = \overline{f(x)}$, $\pi_{0}(g)(\overline{x}) = \overline{g(x)}.$ So, we wish to show that $\overline{f(x)} = \overline{g(x)}$ for any $x \in X$, for which it suffices to find a path connecting $f(x)$ and $g(x)$ in $Y$ for any $x \in X$.

Fix an arbitrary $x_{0} \in X,$ and let $h_{x_{0}} \colon I \to Y$ be defined by $h_{x_{0}}(t) = H(x_{0}, t).$ Check for yourself that $h_{x_{0}}$ is continuous.

Now, note that $$h_{x_{0}}(0) = H(x_{0}, 0) = f(x_{0}),$$ $$h_{x_{0}}(1) = H(x_{0}, 1) = g(x_{0}),$$ so $h_{x_{0}}$ is a path joining $f(x_{0})$ and $g(x_{0})$ in $Y$.