By definition I know that a space $X$ is contractible if $X \sim \{x_0\}$, i.e. homotopic to a point. If I want to show that this implies $id_{X} \sim \text{constant map}$ I am doing this:
By definition of contractible, there exist $f: X \rightarrow \{x_0\}$ and $g: \{x_0\} \rightarrow X$ continuous with $f \circ g \sim id_{x_0}$ and $g \circ f \sim id_{X}$. Now I can deduce that $g \circ f = \text{const}$. Hence $id_{X} \sim \text{const}$.
My question is: when I write $g \circ f = \text{const}$, is it possible that the constant map is the map that sends all $x$ to a point $x_1$, whith $x_1$ a priori different than the point $x_0$ (the $x_0$ present in the definition of $X$ contractible), correct?
Because my reasoning is the following: it is true that the function $g \circ f$ is constant (since $f$ maps all $X$ to the singleton), but nobody tells me that $g$ is the identity: a priori it can be any map sending $x_0$ to any other different $x_1 \in X$, still being continuous as constant map. So if one writes $$X \sim \{x_0\} \iff id_{X} \sim \text{const}_{x_0},$$ where I force $x_0$ to be the same point, this equivalence would be false right?
In other terms, would I be allowed to say $$X \sim \{x_0\} \iff id_{X} \sim \text{const}_{x_0},$$ where I force $x_0$ to be the same point for any other reason that I do not see now? Thanks!
Terminology correction: maps are homotopic, spaces are homotopy-equivalent.
Note that $\{x_0\}$ in the definition of contractibility could be any singleton. There is no reason for $x_0$ to even be a point of $X$. If $x_0$ is not a point of $X$, the constant map with value $x_0$ is not a map $X\rightarrow X$, so asking it to be homotopic to the identity of $X$ simply does not make sense as a statement.
If we write $x_1=g(x_0)\in X$, then you have correctly observed that $g\circ f$ is homotopic to the identity of $X$ on one hand, yet on the other hand is the constant map $X\rightarrow X$ with value $x_1$, let me call that $c_{x_1}\colon X\rightarrow X$. This shows that $id_X\sim c_{x_1}$ is nullhomotopic (which, by definition, means homotopic to a constant map). However, $X$ contractible implies that it is path-connected (why?), and if $\gamma\colon I\rightarrow X$ is a path s.t. $\gamma(0)=x_1$ and $\gamma(1)=x_2$, then $H\colon X\times I\rightarrow X,\,(x,t)\mapsto\gamma(t)$ is a homotopy $c_{x_1}\sim c_{x_2}$. Thus, it turns out that $id_X\sim c_x$ for any $x\in X$.