M. SHTAN’KO

CONTINUA POSSESSING THE FIXED-POINT PROPERTY

(Presented by Academician P. S. Aleksandrov, 23 X 1963)

1. The problem, posed by P. S. Aleksandrov, of the existence of a fixed point under an arbitrary continuous mapping into itself of an acyclic one-dimensional continuum remains unsolved. The strongest result in this direction is the theorem on the existence of a fixed point under an arbitrary continuous self-mapping of every snake-like continuum (O. V. Lokutsievskii (¹), Hamilton (²)) and the theorem on the existence of a fixed point under an arbitrary continuous self-mapping of a linearly connected, one-dimensional, acyclic continuum, proved by K. Borsuk in 1954 (³).

Here theorems on the existence of a fixed point are proved for certain one-dimensional acyclic continua.

1. Theorem 1. If a continuum \(K\) can be \(\varepsilon\)-mapped onto a finite tree \(T_\varepsilon\), \(g_\varepsilon : K \to T_\varepsilon\), in such a way that the inverse images of the branching points of the tree are connected, then the continuum \(K\) possesses the fixed-point property.

The proof is based on the following lemmas.

Lemma 1. Let \(g_\varepsilon : K \to T_\varepsilon\) be an arbitrary \(\varepsilon\)-mapping of the continuum \(K\) onto a finite tree \(T_\varepsilon\), and let \(\widehat{pq}\) be an arc of the tree \(T_\varepsilon\) such that inside \(\widehat{pq}\), i.e. in \(\widehat{pq}\setminus(p\cup q)\), there are no branching points of the tree \(T_\varepsilon\). Then in the compactum
\[ C=g_\varepsilon^{-1}(\widehat{pq}) \]
there exists a continuum \(A\) such that \(g_\varepsilon(A)=\widehat{pq}\).

The proof of Lemma 1 is carried out by elementary set-theoretic methods.

Let \(\operatorname{komp}_x X\) denote the component of the point \(x\) in the set \(X\).

Definition 1. We say that a point \(x\) belonging to the continuum \(A\) of the preceding lemma goes to the left under the mapping \(f\), if one of the following two conditions is satisfied:

1) \(g_\varepsilon(x)\ne q\) and
\[ g_\varepsilon f(x)\in T_\varepsilon\setminus \operatorname{komp}_q\bigl(T_\varepsilon\setminus g_\varepsilon(x)\bigr); \]

2) \(g_\varepsilon(x)=q\) and
\[ g_\varepsilon f(x)\in \overline{\operatorname{komp}_p\bigl(T_\varepsilon\setminus g_\varepsilon(x)\bigr)}. \]

The point \(x\) goes to the right if in 1) and 2) one replaces \(q\) by \(p\).

Lemma 2. If in the continuum \(A\) of Lemma 1 there exist two points \(x\) and \(y\) such that under the continuous mapping \(f\) the point \(x\) goes to the left and the point \(y\) goes to the right, then in the continuum \(A\) there exists a point \(a\in A\) such that
\[ g_\varepsilon(a)=g_\varepsilon f(a). \]

Proof. Every point \(a\in A\) goes either to the left or to the right, and the set of points going to the left (to the right) is closed. The continuum \(A\) is the sum of two nonempty closed sets, and for every point \(a\) in their intersection
\[ g_\varepsilon(a)=g_\varepsilon f(a). \]

To prove Theorem 1 it is enough to show that there is a continuum \(A\) satisfying the conditions of Lemmas 1 and 2. This can be done by using the connectedness property of the inverse images of the branching points of the tree \(T_\varepsilon\). Therefore

there exists a point \(a\in K\) such that \(g_\varepsilon(a)=g_\varepsilon f(a)\), and since \(g_\varepsilon\) is an \(\varepsilon\)-mapping, \(\rho(a,f(a))<\varepsilon\). In view of the arbitrariness of \(\varepsilon\), the assertion of Theorem 1 follows.

From Theorem 1 the previously known theorems easily follow:

Corollary 1. Every snakelike continuum has the fixed-point property.

Corollary 2. A tree (a locally connected one-dimensional continuum containing no circle) has the fixed-point property, since every tree can be monotonically \(\varepsilon\)-mapped onto a finite tree.

1. Let the continuum \(K\) be the sum of two continua: \(K=K_1\cup K_2\), where \(K_1\cap K_2\) is a continuum and each of the continua \(K_1,K_2\) has the fixed-point property. It is unknown whether, in this case, the continuum \(K\) has the fixed-point property.

If \(K_1\cap K_2\) is a point, then the problem is solved simply. Let \(f:K\to K\) be an arbitrary continuous mapping of the continuum \(K\) into itself. Let \(K_1\cap K_2=x\); if \(f(x)=x\), then a fixed point exists; suppose \(f(x)\in K_2\setminus x\), and let \(\varphi:K\to K\) be a mapping identical on \(K_2\) and sending each point \(y\in K_1\setminus x\) to the point \(x\). Then, for the mapping \(\varphi f:K_2\to K_2\), by hypothesis there exists a point \(z\in K_2\) such that \(\varphi f(z)=z\). But \(z\ne x\), since \(\varphi f(x)=f(x)\ne x\); hence \(z\in K_2\setminus x\) and \(\varphi f(z)=f(z)=z\).

Theorem 2. If the continuum \(K\) is the sum of two of its proper subcontinua: \(K=K_1\cup K_2\) and \(K_1\cap K_2=T\) is a tree, and if both continua \(K_1\) and \(K_2\) are one-dimensional and each of them has the fixed-point property, then the continuum \(K\) has the fixed-point property.

Proof. If the continuum \(K\) contained a circle, then the circle would lie in one of the continua \(K_i\), \(i=1,2\); but then the continuum \(K_i\) would not have the fixed-point property, and therefore \(K\) contains no circle.

Consider two cases:

1) \(f(T)\cap T=\Lambda\); \(f(T)\subset K_2\setminus T\). Every tree \(T\) is an absolute retract; hence there exists a continuous mapping \(\varphi:K_1\to T\) (a retraction), identical on \(T\). Extend the mapping \(\varphi\) to the whole continuum \(K\), making it identical on \(K_2\). Then the mapping \(\varphi f:K_2\to K_2\) is a mapping of the continuum \(K_2\) into itself. By hypothesis there exists a fixed point for this mapping: \(x\in K_2\), \(\varphi f(x)=x\). The point \(x\in T\), since for \(x'\in T\), \(f(x')\in K_2\setminus T\) and \(\varphi f(x')=f(x')\ne x'\). Let \(f(x)=y\). If \(y\in K_1\), then \(\varphi(y)\in T\) and, consequently, \(\varphi(y)\ne x\). If, however, \(y\in K_2\setminus K_1\), then \(\varphi(y)=y\) and, consequently, \(\varphi f(x)=f(x)=x\); but then the point \(x\) will be fixed also under the mapping \(f:K\to K\).

2) \(f(T)\cap T\ne \Lambda\). We shall give here only the idea of the proof. \(f(T)\) is a one-dimensional locally connected continuum containing no circle; hence \(f(T)\) is a tree. \(f(T)\cap T=B\) is connected, since \(K\) contains no circle. Define a mapping \(\psi:f(T)\to B\) (a contraction) as follows: \(\psi f(x)=f(x)\) if \(f(x)\in B\), and each component \(f(T)\setminus B\) is mapped to the point
\[ x=B\cap \overline{\operatorname{comp}(f(T)\setminus B)}. \]
The mapping \(\psi f:T\to B\) is a mapping of the tree \(T\) into itself.

It is first proved that if, under the mapping \(\psi f:T\to B\), there is more than one fixed point, then the mapping \(f:K\to K\) also has a fixed point. Suppose the mapping \(\psi f:T\to B\) has only one fixed point \(m\): \(\psi f(m)=m\). If \(f(m)\in T\), then \(f(m)\in B\) and \(\psi f(m)=f(m)=m\).

Now suppose that \(f(m)\in K_2\setminus T\). Define the following mapping:

1) \(\psi':(f(T)\cap K_1)\to B\) is a contraction and is defined analogously to \(\psi:f(T)\to B\).

2) \(\psi'\) is a homeomorphism on the set \(K_1\setminus (f(T)\cap K_1)\).

3) \(\psi'\) is the identity mapping on \(K_2\); then \(\psi' : K \to K_2 \cup C\), where \(C = \psi'(K_1)\).

Define the mapping (retraction) \(\varphi : (K_2 \cap C) \to K_2\), identical on \(K_2\) and such that \(\varphi(C) = T\).

Consider the continuous mapping \(\varphi\psi'f : K_2 \to K_2\); by the hypothesis, it has a fixed point \(\varphi\psi'f(x) = x\). This point is also fixed under the mapping \(f\): \(f(x) = x\).

Received
2 VII 1963

REFERENCES

1. O. V. Dokuchaevskii, UMN, 12, No. 3, 171 (1957).
2. O. H. Hamilton, Proc. Am. Math. Soc., 2, 173 (1951).
3. K. Borsuk, Bull. Polish Acad. Sci., Sect. 3, 2, No. 1, 15 (1954).