Let $X$ be a space. Assume that $a\in X$, and the map $c_a:X\to \{a\}$ is homotopic to the identity map $id:X\to X$.
Prove that $X$ is path-connected.
Prove that for all $b\in X$ there is a homotopy from the constant map $c_b:X\to \{b\}$ to $id$.
In 1: I need to show that for any two points $x,y\in X$ there is a continuous map $f:I\to X$ of the unit interval $I=[0,1]$ such that $f(0)=x$ and $f(1)=y$. I thought of applying the def of homotopy, And get that there is a homotopy=continuous map $H=H(x,-): I\to X$ (for a fixed $x\in X$) such that
$H(0)=H(x,0)=c_a(x)=a$ ($\in X$), $H(1)=H(x,1)=x$ ($\in X$)
So, for any two points $x,a \in X$ consider $H$ as the needed map (i.e. $f:=H$).
In 2, I tried to use the def of homotopy and part 1, but how to do so, if for example I define a homotopy $H:X\times I\to X$ as $H(x,t)=t(x-b)+b$ that indeed satisfy $H(x,0)=c_b(x)=b, H(x,1)=x$, however is $+$ have any meaning in a topological space? I think maybe I need to prove here a more general claim..
Hint for 1: Concatenate the paths from $x$ to $a$ with the path from $a$ to $y$. The following equivalence holds in more generality:
Assuming this equivalence is known, then you are done with part $a$. If not, this can be proven using the same hint above.
Hint for 2: Since $X$ is path connected, there is a path between $b=f(0)$ and $a=f(1)$. The map $G:X\times [0,1]\to X$ given by $(x,t)\mapsto f(t)$ is a homotopy from $c_b$ and $c_a$. Try to concatenate $G$ with $H$ by performing $G$ first, then $H$, where $H$ is the homotopy from $c_a$ to $\text{id}_X$. This concatenation of homotopies will result in a homotopy between $c_b$ and $\text{id}_X$.