As stated in the title, I am working on this question
Show that if $f:X\longrightarrow Y$ is a homotopy equivalence, then it is unique up to homotopy.
I have some attempt but I got stuck and now even doubt the validity of this problem…
My attempt:
Let $f_{1}:X\longrightarrow Y$ be a homotopy equivalence with the homotopy inverse $f_{2}:Y\longrightarrow X$. Let $g_{1}:X\longrightarrow Y$ be another homotopy equivalence with the homotopy inverse $g_{2}:Y\longrightarrow X$.
Then we show that $f_{1}\sim g_{1}$ and $f_{2}\sim g_{2}$
By the definition of homotopy equivalence, we have $$f_{2}\circ f_{1}\sim Id_{X}\sim g_{2}\circ g_{1},$$ and $$f_{1}\circ f_{2}\sim Id_{Y}\sim g_{1}\circ g_{2}.$$
Since homotopy is an equivalence relation, we have $f_{2}\circ f_{1}\sim g_{2}\circ g_{1}$ and $f_{1}\circ f_{2}\sim g_{1}\circ g_{2}$.
Then I got stuck since I could not use the information in hand to construct a homotopy between $f_{1}$ and $g_{1}$.
Is this question even correct? I think if $f$ and the corresponding homotopy inverse preserves the base point, i.e. $f_{1}(x_{0})=y_{0}$ and $f_{2}(y_{0})=x_{0}$, I may be able to generate a proof using the fact that the induced homomorphisms $(f_{1})_{*}$, $(f_{2})_{*}$, $(g_{1})_{*}$ and $(g_{2})_{*}$ are isomorphism of fundamental group.
However, for now I don't know what to do.
Any idea or counter-example that this exercise is wrong? Thank you!
Edit:
So given the argument given by Max, the above statement is incorrect. This edit is to give a formal proof of the counter-example, which basically follows Max's idea, I just added more details from the communications we made in the comment section. In particular, I showed that $Id$ and $-1$ cannot be homotopic with more details.
Consider the $Id_{\mathbb{S}^{1}}:\mathbb{S}^{1}\longrightarrow\mathbb{S}^{1}$ as opposed to $-1:\mathbb{S}^{1}\longrightarrow\mathbb{S}^{1}$, for instance $f(t)=e^{-2\pi it}$ on $t\in [0,1]$. They are both homotopy equivalences since in fact they are both homeomorphisms, but they are not homotopic to each other.
Indeed, for brevity denote $f:=Id_{\mathbb{S}^{1}}$ and $g:=-1$, and then consider $1\in\mathbb{S}^{1}$ be the base point. Then, we have $f(1)=1$ but $g(1)=-1$. Thus, we have the maps $$f:(\mathbb{S}^{1}, 1)\longrightarrow(\mathbb{S}^{1},1)\ \text{and}\ g:(\mathbb{S}^{1}, 1)\longrightarrow(\mathbb{S}^{1}, -1),$$ and thus we have the induced homomorphism $$f_{*}:\pi_{1}(\mathbb{S}^{1}, 1)\longrightarrow\pi_{1}(\mathbb{S}^{1}, 1)\ \text{and}\ g_{*}:\pi_{1}(\mathbb{S}^{1},1)\longrightarrow\pi_{1}(\mathbb{S}^{1},-1).$$
Then, if $f$ and $g$ were homotopic, then there would be a path $\alpha$ in $\mathbb{S}^{1}$ from $1$ to $-1$, such that $$g_{*}=\Phi_{[\alpha]}\circ f_{*},$$ where $\Phi_{[\alpha]}:\pi_{1}(\mathbb{S}^{1},1)\longrightarrow\pi_{1}(\mathbb{S}^{1}, -1)$ is the know isomorphism defined by $[h]\mapsto[\alpha^{-1}*f*\alpha]=[\alpha]^{-1}*[h]*[\alpha]$.
Thus, let $[h]\in\pi_{1}(\mathbb{S}^{1}, 1)$, we have $$[g\circ h]=g_{*}([h])=\Phi_{[\alpha]}(f_{*}([h]))=[\alpha]^{-1}*[f\circ h]*[\alpha],$$ but don't froget $\pi_{1}(\mathbb{S}^{1}, x_{0})\cong\mathbb{Z}$ which is abelian given any base point $x_{0}$, and thus we can cancel $[\alpha]^{-1}$ with $[\alpha]$ on the RHS of the above equation, and thus we have $$[g\circ h]=[f\circ h]$$ for all $[h]\in \pi_{1}(\mathbb{S}^{1},1)$.
In other word, $g_{*}$ and $f_{*}$ agree on $\pi_{1}(\mathbb{S}^{1}, 1)$, but look at what they really are! $g_{*}=(-1)_{\pi_{1}(\mathbb{S}^{1},1 )}$ and $f_{*}=Id_{\pi_{1}(\mathbb{S}^{1},1)}$, which means that $f_{*}$ is the generator of this group (since $Id$ is the generator of this group), and $g_{*}$ is the inverse of this group (since $-1$ is the inverse of this group).
But this group is isomorphic to $\mathbb{Z}$ in which no one is equal to its inverse, a contradiction.
Thus, $f$ and $g$ are not homotopic to each other.
Best Answer
The question is definitely not correct. Take for instance the identity $S^1 \to S^1$ as opposed to $-1 : S^1\to S^1$ ($-1$ is repeesented by $f(t) = e^{-2i\pi t}$ for instance, defined on $I=[0,1]$)
They are both homotopy equivalences (and in fact, both homeomorphisms), but they're not homotopic (if you know what $\pi_1(S^1)$ is, it follows by some usual property relating pointed homotopy classes and unpointed homotopy classes - of course if you're interested in pointed homotopy equivalences, and pointed homotopies, this is just $\pi_1(S^1)$)