Is there an $X$ and a non-surjective continuous function $f:X\to X$ such that $f\simeq {\rm id}_X$ but $f(X)$ and $X$ are not homotopy equivalent?

Question

Let $X\subset \mathbb{Z}^n$ be a finite set (vertex set) and $1\leq u\leq n$. For $x=(x_1 ,\ldots ,x_n)\neq y=(y_1 ,\ldots ,y_n) \in X$, we define the adjacency $\kappa_u$ on $X$ as follows: $x\sim_{\kappa_u}y$ if and only if (1): for at most $u$ indices $i$ we have $|x_i -y_i|=1$, and (2) for all indices $j$, $|x_j -y_j|\neq 1$ implies $x_j =y_j$. By $[0,m]_{\mathbb{Z}}$ I mean $0\sim_{\kappa_1}1\sim_{\kappa_2}\sim \cdots \sim_{\kappa_1}m$. $f:X\to X$ is continuous if $x\sim_{\kappa_u}y$ implies $f(x)\sim_{\kappa_u}f(y)$. $f,g:X\to X$ are homotopic if there is a function $h:X\times [0,m]_{\mathbb{Z}}\to X$ such that: (1) for all $x\in X$, $h(x,0)=f(x)$ and $h(x,m)=g(x)$; (2) for all $x\in X$, $h_x :[0,m]_{\mathbb{Z}}\to X$ by $h_x (t)=h(x,t)$ is continuous, and (3) for all $t\in [0,m]_{\mathbb{Z}}$, $h_t :X\to X$ $h_t (x)=h(x,t)$ is continuous. If there exist continuous $f:X\to Y$ and $g:Y\to X$ such that $gf$ is homotopic to ${\rm id}_X$ and $fg$ is homotopic to ${\rm id}_Y$, we say $X$ and $Y$ are homotopy equivalent.

My question: is there a finie $X$ and a non-surjective continuous function $f:X\to X$ such that $f\simeq {\rm id}_X$ but $f(X)$ and $X$ are not homotopy equivalent?

Answer 1

I’m not familiar with this particular model of spaces and homotopies. The following is a possible example, depending on an assumption which seems likely, and I guess should be well-known in this setting if it’s true: that for sufficiently large $N \gg u$, the boundary of an $N \times N$ square in the “plane” $\newcommand{\Z}{\mathbb{Z}} \Z^2$ is not homotopy equivalent to a point. This seems geometrically reasonable: it looks like it should be a model of the circle.

In topological spaces, a natural example is $f : \newcommand{\R}{\mathbb{R}}\R^2 \to \R^2$, $f(x,y) = (\cos x, \sin x)$, or more generally, its restriction to any large enough square $[0,a]_\R^2$ — that is, take a plane square, project it to a line, and use that line to parametrise a circle in the plane. The image is the circle, which is not contractible, so not homotopy equivalent to the plane. But the map is homotopic to the identity just by linear interpolation, or more abstractly, since the original plane region is contractible.

Now we can hopefully adapt this to your setting. Given $u$, take $N \gg u$, and take your space $X$ to be $[0,4N]_\Z^n$. Now take the map $f : [0,4N]_\Z^n \to [0,4N]_\Z^n$ given by composing the first projection $[0,4N]_\Z^n \to [0,4N]_\Z$ with the map $[0,4N]_\Z \to [0,4N]_\Z^2$ that uses the line to trace out the perimeter of some $N \times N$ square, and then the injection from $[0,4N]_\Z^2 \to [0,4N]_\Z^n$ with some arbitrary fixed value on the remaining $n-2$ co-ordinates. Now $f$ is certainly homotopic to $\mathrm{id}_X$, since each is homotopic to a constant map. But its image is the boundary of the $N \times N$ square. So assuming the boundary of a large enough square in the plane is not contractible, this example gives what you ask for.