Maps That Undo One Another Up to Homotopy
A homeomorphism has an inverse that recovers every point exactly. Homotopy equivalence asks for less: continuous maps in both directions whose composites can be continuously deformed to the identity maps. The two composite conditions are both essential. We will state the definition, verify it through explicit examples, and prove that homotopy equivalent spaces have the same path-connectedness behavior.
Throughout, all spaces are topological spaces, and products have the product topology. Write \(I=[0,1]\) with its standard topology. A homotopy from \(f:X\to Y\) to \(f_1:X\to Y\) is a continuous map \(H:X\times I\to Y\) such that \(H(x,0)=f(x)\) and \(H(x,1)=f_1(x)\) for every \(x\in X\). We write \(f\simeq f_1\). Reversing the parameter gives a homotopy in the opposite direction: if \(H\) goes from \(f\) to \(f_1\), then \((x,t)\mapsto H(x,1-t)\) goes from \(f_1\) to \(f\).
The order of the maps matters when writing the conditions: \(g\circ f\) is a map from \(X\) to \(X\), so it must be compared with \(\operatorname{id}_X\); \(f\circ g\) is a map from \(Y\) to \(Y\), so it must be compared with \(\operatorname{id}_Y\). A homotopy inverse is not required to undo \(f\) point by point. Instead, each composite must be homotopic to the appropriate identity.
First Examples: A Point and the Real Line
Worked Example: The Real Line and a One-Point Space
Let \(P=\{p\}\) have its unique topology, and give \(\mathbb{R}\) its standard topology. Define \(f:\mathbb{R}\to P\) by \(f(x)=p\), and \(g:P\to\mathbb{R}\) by \(g(p)=0\). Both maps are continuous: \(f\) is constant, and every map from a one-point space is continuous.
The composite on \(P\) satisfies \((f\circ g)(p)=f(0)=p\), so \(f\circ g=\operatorname{id}_P\). On \(\mathbb{R}\), \((g\circ f)(x)=g(p)=0\). Thus \(g\circ f\) is the constant-zero map, not the identity map.
Define \(H:\mathbb{R}\times I\to\mathbb{R}\) by \(H(x,t)=(1-t)x\). This is continuous because it is given by sums and products of continuous real-valued coordinate functions. Its endpoints are \(H(x,0)=x\) and \(H(x,1)=0\): it goes from \(\operatorname{id}_{\mathbb{R}}\) to \(g\circ f\). To obtain the endpoint order required by \(g\circ f\simeq\operatorname{id}_{\mathbb{R}}\), reverse the parameter: \(\widetilde H(x,t)=H(x,1-t)=tx\). Then \(\widetilde H(x,0)=0=(g\circ f)(x)\) and \(\widetilde H(x,1)=x=\operatorname{id}_{\mathbb{R}}(x)\). Hence both composite conditions hold, and \(f\) is a homotopy equivalence.
Check: \(f\circ g\) is exactly the identity on \(P\), while the reversed homotopy has endpoints \(g\circ f\) first and \(\operatorname{id}_{\mathbb{R}}\) second, as required.
This example also illustrates why one must distinguish a homotopy from an equality. The composite \(g\circ f\) sends every real number to \(0\), so it is not equal to \(\operatorname{id}_{\mathbb{R}}\). The explicit homotopy, rather than pointwise equality, verifies the condition.
Worked Example: The Plane and the Real Line
Give \(\mathbb{R}^2\) and \(\mathbb{R}\) their standard topologies. Define \(f:\mathbb{R}^2\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}^2\) by \(f(x,y)=x\) and \(g(x)=(x,0)\). These coordinate maps are continuous.
For \(x\in\mathbb{R}\), \((f\circ g)(x)=f(x,0)=x\), so \(f\circ g=\operatorname{id}_{\mathbb{R}}\). For \((x,y)\in\mathbb{R}^2\), \((g\circ f)(x,y)=g(x)=(x,0)\). This composite is not the identity at points with \(y\ne0\).
Define \(H:\mathbb{R}^2\times I\to\mathbb{R}^2\) by \(H((x,y),t)=(x,ty)\). Its coordinate functions are continuous, so \(H\) is continuous. At \(t=0\), \(H((x,y),0)=(x,0)=(g\circ f)(x,y)\); at \(t=1\), \(H((x,y),1)=(x,y)=\operatorname{id}_{\mathbb{R}^2}(x,y)\). Therefore \(g\circ f\simeq\operatorname{id}_{\mathbb{R}^2}\), and the exact identity \(f\circ g=\operatorname{id}_{\mathbb{R}}\) supplies the other condition. Thus \(f\) is a homotopy equivalence with homotopy inverse \(g\).
Check: at \(t=0\) the second coordinate is \(0\), and at \(t=1\) it is \(y\); the first coordinate remains \(x\) throughout. These are precisely the two required endpoint maps.
Homotopies Depend on the Topology
Worked Example: A Two-Point Indiscrete Space
Let \(X=\{a,b\}\) have the indiscrete topology \(\{\varnothing,X\}\), and let \(P=\{p\}\) have its unique topology. Define \(f:X\to P\) by \(f(a)=f(b)=p\), and \(g:P\to X\) by \(g(p)=a\). Both are continuous. The composite \(f\circ g\) is \(\operatorname{id}_P\), while \(g\circ f\) is the constant map \(c_a:X\to X\) with value \(a\).
Define \(H:X\times I\to X\) by \(H(x,0)=a\) and \(H(x,t)=x\) for \(t>0\). This set map need not have any special continuity behavior in its formula, but it is continuous: the only open subsets of its codomain \(X\) are \(\varnothing\) and \(X\), whose preimages are respectively \(\varnothing\) and all of \(X\times I\), both open. Also \(H(x,0)=c_a(x)\) and \(H(x,1)=x\), so \(c_a\simeq\operatorname{id}_X\). Thus \(f\) and \(g\) satisfy both composite conditions.
Check: The homotopy’s endpoint at \(t=0\) is the constant map to \(a\); its endpoint at \(t=1\) is the identity. The continuity argument uses the indiscrete topology on the target, not a claim that the formula is continuous as a map to a discrete space.
The example is useful because it separates the definition from geometric intuition about a deformation. Continuity is determined by the topologies on the domain and codomain. A map into an indiscrete space is always continuous, since there are only two open sets to check.
Path-Connectedness Is Preserved
A space \(X\) is path-connected if for every \(x_0,x_1\in X\) there is a continuous path \(\alpha:I\to X\) with \(\alpha(0)=x_0\) and \(\alpha(1)=x_1\). The following result gives a useful necessary consequence of homotopy equivalence.
Proof. Let \(g:Y\to X\) be a homotopy inverse, and choose homotopies \(K:X\times I\to X\) from \(g\circ f\) to \(\operatorname{id}_X\), and \(L:Y\times I\to Y\) from \(f\circ g\) to \(\operatorname{id}_Y\).
First suppose \(X\) is path-connected. For any \(y_0,y_1\in Y\), choose \(x_0,x_1\in X\) with \(g(y_0)\) and \(g(y_1)\) as endpoints of a path \(\alpha:I\to X\); such a path exists by path-connectedness. Then \(f\circ\alpha\) is a path in \(Y\) from \(f(g(y_0))\) to \(f(g(y_1))\). For each \(i\in\{0,1\}\), the map \(t\mapsto L(y_i,t)\) is a path from \(f(g(y_i))\) to \(y_i\). Reverse the first such path to go from \(y_0\) to \(f(g(y_0))\), follow \(f\circ\alpha\), and then follow the path \(t\mapsto L(y_1,t)\). Concatenating these three paths gives a path from \(y_0\) to \(y_1\). Thus \(Y\) is path-connected.
Now suppose \(Y\) is path-connected, and take any \(x_0,x_1\in X\). Choose a path \(\beta:I\to Y\) from \(f(x_0)\) to \(f(x_1)\). The composite \(g\circ\beta\) is a path from \(g(f(x_0))\) to \(g(f(x_1))\). For each \(i\in\{0,1\}\), \(t\mapsto K(x_i,t)\) goes from \(g(f(x_i))\) to \(x_i\). Reverse the path for \(i=0\), follow \(g\circ\beta\), and then follow the path for \(i=1\). The concatenation joins \(x_0\) to \(x_1\), so \(X\) is path-connected. The path reversals and concatenations used here are continuous, by the standard path operations. \(\square\)
The statement is an if-and-only-if: each direction uses the homotopy inverse and the corresponding composite homotopy. A continuous map in just one direction would not by itself provide the paths needed to transfer path-connectedness back.
A Common Pitfall: Checking Only One Composite
Having one composite homotopic to an identity does not establish a homotopy equivalence. To see this, let \(X=\{p\}\) and let \(Y=\{a,b\}\) have the discrete topology, whose every subset is open. Define \(f:X\to Y\) by \(f(p)=a\), and let \(g:Y\to X\) be the unique map. Both maps are continuous, and \(g\circ f=\operatorname{id}_X\) exactly. But \((f\circ g)(a)=a\) and \((f\circ g)(b)=a\), so \(f\circ g\) is constant and differs from \(\operatorname{id}_Y\).
There is no homotopy from this constant map to \(\operatorname{id}_Y\). If such a homotopy existed, fixing \(b\) would give a continuous path \(t\mapsto H(b,t)\) in \(Y\) from \(a\) to \(b\). A continuous path into a discrete two-point space cannot have distinct endpoints: regarding its values \(a,b\) as real numbers \(0,1\), the resulting real-valued function is continuous, so the intermediate value theorem would force it to take the value \(1/2\), which is not in \(\{0,1\}\). Therefore the second composite condition fails, and \(f\) is not a homotopy equivalence.
When testing a proposed homotopy equivalence, calculate both composites before constructing homotopies. Record their domains and codomains, compare each with the identity on that same space, and verify the endpoints of each homotopy in the stated order. Reversing a homotopy’s parameter is always available when the endpoints occur in the opposite order.
Check Your Understanding
For each question, keep track of the domain and codomain of both composites.
- In the real-line example, what is \(g\circ f\), and which parameter-reversed homotopy has it as its initial map?
- For the maps between \(\mathbb{R}^2\) and \(\mathbb{R}\), write both composites and identify which one is exactly an identity.
- Why is every map from the two-point indiscrete space to itself continuous? Does that fact alone establish a homotopy between two specified maps?
- In the discrete two-point pitfall, which composite condition holds exactly, and why does the other one fail?
- Use the path-connectedness theorem to explain why a homotopy equivalence from a path-connected space cannot have a non-path-connected target.