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A GROUP OF AUTOMORPHISMS OF THE HOMOTOPY GROUPS

HIROSHI UEHARA

It is well known that the fundamental group $\pi_1(X)$ of an arcwise connected topological space $X$ operates on the $n$-th homotopy group $\pi_n(X)$ of $X$ as a group of automorphisms. In this paper I intend to construct geometrically a group $\mathcal{H}(X)$ of automorphisms of $\pi_n(X)$, for every integer $n \ge 1$, which includes a normal subgroup isomorphic to $\pi_1(X)$, so that the factor group of $\mathcal{H}(X)$ by $\pi_1(X)$ is completely determined by some invariant $\mathcal{L}(X)$ of the space $X$. The complete analysis of the operation of the group on $\pi_n(X)$ is given in §3, §4, and §5.

Throughout the whole paper, $X$ denotes an arcwise connected topological space which has such suitable homotopy extension properties as a polyhedron does, and all mappings are continuous transformations.

§1. Definition of the group $\mathcal{H}(X)$.

Let $x_0$ be an arbitrary point of the space $X$, and $\Omega$ a collection $\mathcal{X}^*(x_0, x_0)$ of all the mappings that transform $X$ into $X$ and $x_0$ into $x_0$. For two maps $a, b \in \Omega$, $a$ is said to be homotopic to $b$ (in notation : $a \sim b$) if there exists a homotopy $h_t \in \Omega$ (for $1 \le t \le 0$) such that $h_0 = a$ and $h_1 = b$. A mapping $a \in \Omega$ is called to have a (two sided) homotopy inverse, if there is a map $\varphi \in \Omega$ such that $\alpha\varphi \sim 1$ and $\varphi\alpha \sim 1$, where 1 denotes the identity transformation of $X$ onto itself. Let $\Omega^*$ be the collection of all the mappings belonging to $\Omega$, each of which has a homotopy inverse.

Now let $X \times I$ be the topological product of $X$ and the line segment $I$ between 0 and 1, and let us consider the totality $U$ of the mappings $\vartheta : X \times I \rightarrow X$ which satisfy the following conditions :

For two maps $\theta, \theta' \in U$, $\theta$ is homotopic to $\theta'$ (notation : $\theta \sim \theta'$) if there exists a homotopy $h_t : X \times I \to X$ (for $1 \le t \le 0$) such that

Received Oct. 25, 1950.

I should like to express my sincere gratitude for the courtesies extended to me by Professor S. T. Hu. This paper is inspired by his paper, "On the Whitehead group of automorphisms of the relative homotopy groups." ---PAGE_BREAK---

It is easily verified that this relation is an equivalent relation, and therefore $U$ is divided into equivalent classes in this sense.

We shall denote by $[\theta]$ the class containing $\theta$. For $\theta \in U$ we construct a mapping $\sigma_0 \in U$ as follows: a mapping $\bar{\sigma}_\theta$ which is defined continuously on the set $((X \times 0) \uplus (x_0 \times I))$ such that $\bar{\sigma}_\theta(x, 0) = x$ and $\bar{\sigma}_\theta(x_0, t) = \theta(x_0, t)$, can be extended to a mapping $\sigma_0 \in U$, provided that ${x_0}$ has a homotopy extension property in $X$ relative to $X$. The extended mapping is, of course, not unique but the homotopy class containing $\sigma_0$ is uniquely determined if the set $((x_0 \times I) \uplus (X \times 0) \uplus (X \times 1))$ has a homotopy extension property in $X \times I$ relative to $X$; another arbitrarily extended map $\sigma'_0$ is homotopic to $\sigma_0$. Now two maps $\theta_1, \theta_2 \in U$ are 'multiplied' together by the rule,

where $\rho(x, t) = \theta_2(\theta_1(x, t), 0)$. Then we have

LEMMA 1.1 $\theta_1 \times \theta_2$ is again a member of the collection $U$.

Proof. Let $a_1(x) = \theta_1(x, 0)$, $a_2(x) = \theta_2(x, 0)$, then both $a_1$ and $a_2$ belong to $\Omega^*$, so that $a_1$ and $a_2$ have homotopy inverses $\varphi_1, \varphi_2$ respectively. From the considerations that $\varphi_1\varphi_2$ is a homotopy inverse of $\omega a_1$ and that $\theta_1 \times \theta_2(x, 0) = \rho(x, 0) = \theta_2(\theta_1(x, 0), 0) = \theta_2(a_1(x), 0) = a_2(a_1(x))$, we have $\theta_1 \times \theta_2 | X \times 0 = \Omega^*$ and therefore the condition (1.1) i) is satisfied. Also we have $\theta_1 \times \theta_2(x_0, 1) = \sigma_{\theta_2}(\rho(x_0, 1), 1) = \sigma_{\theta_2}(x_0, 1) = \theta_2(x_0, 1) = x_0$. This proves the Lemma.

LEMMA 1.2 The class $[\theta_1 \times \theta_2]$ depends only on the classes $[\theta_1]$ and $[\theta_2]$.

Proof. Let $\theta'_1 \in [\theta_1]$ and $\theta'_2 \in [\theta_2]$, then there exist two homotopies $h_s$, $k_s$: $X \times I \rightarrow X$ ($1 \le s \le 0$) such that $h_0 = \theta'_1$, $h_1 = \theta'_2$, $k_0 = \theta'_2$, and $k_1 = \theta'_2$. Putting $\rho_s(x, t) = k_s(h_s(x, t), 0)$, we have

Since $k_s(x_0, 0) = k_s(x_0, 1) = x_0$, we can construct, in virtue of the homotopy extension properties previously mentioned, $\sigma_{k_s} \in U$ ($1 \le s \le 0$), which is also continuous with respect to $\epsilon$, just as in case of $\sigma_\epsilon$. Then clearly we have $\sigma_{k_s}(x, 0) = x$ and $\sigma_{k_s}(x_0, t) = k_s(x_0, t)$ by the construction of the function $\sigma_{k_s}$.

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is obviously continuous and satisfies the conditions (1.2) of the homotopy; as to the condition ii), we have $H_3(x_0, 0) = \rho_5(x_0, 0) = x_0$ from (1.4) ii) and $H_3(x_0, 1) = \sigma_{k_8}(\rho_5(x_0, 1), 1) = \sigma_{k_5}(x_0, 1) = k_5(x_0, 1) = x_0$ from (1.4) iii).

Since (1.2) i) is evidently satisfied from (1.4) i), the lemma has been proved. Thus the multiplication in $U$ induces a multiplication in the set of the homotopy classes ; $[\theta_1] \times [\theta_3] \equiv [\theta_1 \times \theta_2]$.

THEOREM 1. By the multiplication defined above, all the homotopy classes of $U$ constitute a group $\mathfrak{A}(X)$ with $x_0$ as the base point.

Proof. Let us prove that the multiplicatoin is associative. Let $\theta_1, \theta_2, \theta_3 \in U$, then $([\theta_1] \times [\theta_2]) \times [\theta_3]$ and $[\theta_1] \times ([\theta_2] \times [\theta_3])$ are represented by mappings $(\theta_1 \times \theta_2) \times \theta_3$ and $\theta_1 \times (\theta_2 \times \theta_3)$ respectively. By definition

As it is rather difficult to show directly the existence of homotopy between ($\theta_1 \times \theta_2$) $\times$ $\theta_3$ and $\theta_1 \times (\theta_2' \times \theta_3)$, we prove it by making use of the homotopy extension property referred to above. From the relation above we have ($\theta_1 \times \theta_2$) $\times \theta_5 (x, 0) = \theta_3 (\theta_5 (\theta_1 (x, 0), 0), 0) = \theta_1 \times (\theta_2 \times \theta_3) (x, 0)$, and from the property of $\sigma_\theta$ we have

Since $\sigma_{\theta_0 \times \theta_3}(\theta_3(\theta_2(\theta_1(x_0, 1), 0), 0), 2t-1) =: \sigma_{\theta_0 \times \theta_3}(x_0, 2t-1) =$

we have

From (1.6) and (1.7) there exists a homotopy $h(x, s, t)$ defined on ${x_n} \times I^s \times I^t$ ---PAGE_BREAK---

such that

$$h(x_0,0,t)=({\theta }_1\times {\theta }_2)\times {\theta }_3(x_0,t),1\ge t\ge 0,h(x\_0,\; 0,\; t)\; =\; (\backslash theta\_1\; \backslash times\; \backslash theta\_2)\; \backslash times\; \backslash theta\_3(x\_0,\; t),\; \backslash quad\; 1\; \backslash ge\; t\; \backslash ge\; 0,$$

$$h(x_0,1,t)={\theta }_1\times ({\theta }_2\times {\theta }_3)(x_0,t),1\ge t\ge 0,h(x\_0,\; 1,\; t)\; =\; \backslash theta\_1\; \backslash times\; (\backslash theta\_2\; \backslash times\; \backslash theta\_3)(x\_0,\; t),\; \backslash quad\; 1\; \backslash ge\; t\; \backslash ge\; 0,$$

$$h(x_0,s,0)=h(x_0,s,1)=x_0,1\ge s\ge 0.h(x\_0,\; s,\; 0)\; =\; h(x\_0,\; s,\; 1)\; =\; x\_0,\; \backslash quad\; 1\; \backslash ge\; s\; \backslash ge\; 0.$$

and

Moreover putting

$$h(x,0,t)=({\theta }_1\times {\theta }_2)\times {\theta }_3(x,t),x\in X,1\ge t\ge 0,h(x,\; 0,\; t)\; =\; (\backslash theta\_1\; \backslash times\; \backslash theta\_2)\; \backslash times\; \backslash theta\_3\; (x,\; t),\; \backslash quad\; x\; \backslash in\; X,\; \backslash \; 1\; \backslash ge\; t\; \backslash ge\; 0,$$

$$h(x,1,t)={\theta }_1\times ({\theta }_2\times {\theta }_3)(x,t),x\in X,1\ge t\ge 0,h(x,\; 1,\; t)\; =\; \backslash theta\_1\; \backslash times\; (\backslash theta\_2\; \backslash times\; \backslash theta\_3)(x,\; t),\; \backslash quad\; x\; \backslash in\; X,\; \backslash \; 1\; \backslash ge\; t\; \backslash ge\; 0,$$

$$h(x,s,0)={\theta }_3({\theta }_2({\theta }_1(x,0),0),0),x\in X,1\ge s\ge 0,h(x,\; s,\; 0)\; =\; \backslash theta\_3(\backslash theta\_2(\backslash theta\_1(x,\; 0),\; 0),\; 0),\; \backslash quad\; x\; \backslash in\; X,\; \backslash \; 1\; \backslash ge\; s\; \backslash ge\; 0,$$

and

$h$ is defined continuously on the set ${(X \times \frac{s}{I} \times 0) \cup [(x_0 \times \frac{s}{I}) \cup (X \times 0) \cup (X \times 1)] \ \times \frac{t}{I}}$. Thus, if ${(x_0 \times I) \cup (X \times 0) \cup (X \times 1)}$ has a homotopy extension property in $X \times I$ relative to $X$, $h$ can be extended to a mapping $X \times \frac{s}{I} \times \frac{t}{I} \to X$, which gives a homotopy between $(\theta_1 \times \theta_2) \times \theta_3$ and $\theta_1 \times (\theta_2 \times \theta_3)$.

Next we must prove the existence of the unity in $\mathfrak{A}(X)$. Let $\theta_0(x, t) = x$, then clearly $\theta_0 \in U$. For any $\theta \in U$ we have from the definition of multiplication

where $\rho(x, 2t) = \theta_0(\theta(x, 2t), 0) = \theta(x, 2t)$, and $\sigma_{\theta_0}(x, t) = x$ may be assumed. Since $\sigma_{\theta_0}(\rho(x, 1), 2t-1) = \rho(x, 1) = \theta_0(\theta(x, 1), 0) = \theta(x, 1)$ for $1 \le t \le \frac{1}{2}$, we have

Let us define a homotopy $h_s(x, t)$ for $1 \le s \le 0$ as follows;

then $h_s$ satisfies the conditions of the homotopy (1.2), so that $h_0 = \theta \times \theta_0$ and $h_1 = 0$. Thus $\theta_0$ represents the right side unity of the group $\mathfrak{A}(X)$.

Lastly we proceed to show the existence of the inverse element of any element $[\theta] \in \mathfrak{A}(X)$. By the assumption on an element $\theta$ in $U$, we have $\theta |{X \times 0} = 0$, so that $\theta |{X \times 0}$ has a homotopy inverse $\varphi |_{\Omega^*}$. Now we define a mapping $\theta^{-1} : U$ as follows: if we put

then $\theta^{-1}$ can be extended to a map: $X \times I \to X$ because of the homotopy ---PAGE_BREAK---

extension property of {$x_0$}. This extended map $\theta^{-1}$ is shown to represent the inverse of $[\theta]$. Indeed, we have

where $\rho(x, t) = \theta^{-1}(\theta(x, t), 0) = \varphi(\theta(x, t))$, $\sigma_{\theta^{-1}}(x, 0) = x$, and $\sigma_{\theta^{-1}}(x_0, t) = \theta^{-1}(x_0, t) = \varphi(\theta(x_0, 1-t))$. As $\varphi$ is a homotopy inverse of $\theta |{X \times 0}$, and on the other hand $\sigma{\theta^{-1}}|{X_0 \times I}$ represents the inverse element of $[\rho |{X_0 \times I}]$, we have a continuous function $h$ defined on $((X \times \overline{I} \times 0) \cup (X \times 0) \cup (X \times 1)) \cap (X_0 \times \overline{I}) \times \overline{I})$ such that

$$h(x,s,0)=k(x,s),x\in X,s\in \bar{I},h(x,\; s,\; 0)\; =\; k(x,\; s),\; \backslash quad\; x\; \backslash in\; X,\; s\; \backslash in\; \backslash overline\{I\},$$

$$h(x_0,s,t)=l(s,t),s\in \bar{I},t\in \bar{I},h(x\_0,\; s,\; t)\; =\; l(s,\; t),\; \backslash quad\; s\; \backslash in\; \backslash overline\{I\},\; t\; \backslash in\; \backslash overline\{I\},$$

$$h(x,0,t)=\theta \times {\theta }^{-1}(x,t),x\in X,t\in \bar{I},h(x,\; 0,\; t)\; =\; \backslash theta\; \backslash times\; \backslash theta^\{-1\}(x,\; t),\; \backslash quad\; x\; \backslash in\; X,\; t\; \backslash in\; \backslash overline\{I\},$$

$$h(x,1,t)=x,x\in X,t\in \bar{I},h(x,\; 1,\; t)\; =\; x,\; \backslash quad\; x\; \backslash in\; X,\; t\; \backslash in\; \backslash overline\{I\},$$

where $k$ is a homotopy obtained by the relation $\varphi\theta \sim 1$, and $l$ is also a homotopy whose existence is assured by $\rho(x_0, 1-t) = \sigma_{\theta^{-1}}(x_0, t)$. Again, by the aid of a homotopy extension property of $((x_0 \times I) \cup (X \times 0) \cup (X \times 1)))$, $h$ can be extended to a map $X \times I \times I \to X$, which gives a desired homotopy. This completes the proof.

In order to clarify the conditions preassigned to the space $X$ we put down here all the homotopy extension properties assumed in the arguments of the above Theorem;

i) ${x_0}$ has a homotopy extension property in $X$ relative to $X$,

(1.8) ii) ${(x_0 \times I) \cup (X \times 0) \cup (X \times 1)}$ has a homotopy extension property in $X \times I$ relative to $X$.

These assumptions are, of course, satisfied by a polyhedron.

§ 2. A group of automorphisms $\Sigma(X)$ and the structure of $\mathfrak{A}(X)$.

Now we define a group $\Sigma(X)$, which operates on $\pi_n(X)$, as we shall see later, as a group of automorphisms, and study a homomorphism of $\mathfrak{A}(X)$ onto $\Sigma(X)$, the kernel of which is isomorphic to the fundamental group $\pi_1(X)$ of $X$.

Let us define a homotopy concept in $\Omega^*$ in the following sense: we shall write $a \sim b$ for $a, b \in \Omega^*$ if there exists a homotopy $h_t \in \Omega$ ($1 \le t \le 0$) such that $h_0 = a$ and $h_1 = b$. Then $\Omega^*$ is divided into homotopy classes. Let us denote by $\Sigma(X)$ the set of all the homotopy classes. For two maps $a, b \in \Omega^*$ we define $(a \times b)(x) = b(a(x))$ for any $x \in X$. Then $a \times b \in \Omega^*$ because $a \times b \in \Omega$ follows immediately from the definition and, if $\varphi$ and $\psi$ are homotopy inverses of $a$ ---PAGE_BREAK---

and $b$ respectively, $\psi \times \varphi \in \Omega^*$ is a homotopy inverse of $a \times b$. Furthermore, if $a \sim a'$ and $a \sim b'$, $a \times b \sim a' \times b'$. Thus the multiplication in $\Omega^*$ induces a multiplication in $\Sigma(X)$.

THEOREM 2. $\Sigma(X)$ constitutes a group.

Proof. It is evident from the definition of multiplication that the associative law holds. As to the existence of unity, let $E$ be a class containing the identity transformation of $X$, then $E \cdot A = A$ and $A \cdot E = A$ for any $A \in \Sigma(X)$. Lastly for any $A = [a]$ we choose $A^{-1} = [\varphi]$ containing a homotopy inverse $\varphi$ of $a$. Then $AA^{-1} = E$ and $A^{-1}A = E$ is clear from the definition of homotopy inverse.

THEOREM 3. $\Sigma(X)$ operates on the n-th homotopy group $\pi_n(X, x_0)$, for every integer $n \ge 1$, as a group of automorphisms.

Proof. Let $f$ be a representative of an element $\alpha$ of $\pi_n(X)$ and let $\alpha$ be a representative of $A \in \Sigma(X)$. Let us take the mapping $\text{af}: S^n \to X$ as a representative of $A\alpha$. The correspondence $A; \alpha \to A\alpha$ is a transformation of $\pi_n(X)$ into itself because, if $f'$ is another representative of $\alpha$, we have $\text{af} \sim \text{af}'$, and if $\alpha'$ is another representative of $A$, we have also $\text{af} \sim \alpha'f$. Then it is easily proved that this correspondence is an automorphism of $\pi_n(X)$.

Example of $\Sigma(X)$:

Let $X$ be an $n$-sphere $S^n$, then from the concept of Brouwer's degree we have $\Sigma(S^n) = {E = [1], A = [-1]}$ where $E$ is a class containing the identity transformation and $A$ is a class containing a mapping of degree $-1$. Since clearly $A^2 = A \cdot A = E$, the group is a cyclic group of order 2.

Now we intend to define a homomorphism $\varphi$ of $\mathfrak{A}(X)$ onto $\Sigma(X)$. Let $\theta \in U$ be a representative of an element of $\mathfrak{A}(X)$, then $a_\theta = \theta | X \times 0$ represents an element of $\Sigma(X)$. From the homotopy concepts given in §1 and §2, it is obvious that if $\theta \sim \theta'$, we have $a_\theta \sim a_{\theta'}$. By the correspondence $\varphi: [\theta] \to [a_\theta]$ we have the following theorem.

THEOREM 4. $\varphi$ is a homomorphism of $\mathfrak{A}(X)$ onto $\Sigma(X)$, the kernel of which is isomorphic to the fundamental group $\pi_1(X)$.

Proof. For two elements $[\theta_1], [\theta_2] \in \mathfrak{A}(X)$, we have $\varphi([\theta_1]) = [a_{\theta_1}]$ and $\varphi([\theta_2]) = [a_{\theta_2}]$. By definition $\varphi([\theta_1] \times [\theta_2]) = \varphi([\theta_1 \times \theta_2])$ may be represented by a mapping $\theta_1 \times \theta_2 | X \times 0 = \rho(x, 0) = \theta_2(\theta_1(x, 0), 0)$, so that $\theta_1 \times \theta_2 | X \times 0 = a_{\theta_1} \times a_{\theta_2}$. Thus $\varphi([\theta_1] \times [\theta_2]) = \varphi([\theta_1]) \times \varphi([\theta_2])$ is proved. Clearly $\varphi$ is an onto-homomorphism from the definition of the group.

Lastly, in order to complete the proof it is sufficient to prove that the kernel of $\varphi$ is isomorphic to $\pi_1(X)$. If $\varphi([\theta]) = [a_\theta]$ is unity, we may take without loss of generality a representative $\theta$ of $[\theta]$ as follows : ---PAGE_BREAK---

$$(2.1)\begin{array}{l}\text{i)}\theta :X\times I\to X,\\ \text{ii)}\vartheta (x,0)=x,\\ \text{iii)}\vartheta (x_{\theta },1)=x_0,\end{array}\}(2.1)\; \backslash qquad\; \backslash left.\; \backslash begin\{array\}\{l\}\; \backslash text\{i)\}\; \backslash quad\; \backslash theta:\; X\; \backslash times\; I\; \backslash to\; X,\; \backslash \backslash \; \backslash text\{ii)\}\; \backslash quad\; \backslash vartheta(x,\; 0)\; =\; x,\; \backslash \backslash \; \backslash text\{iii)\}\; \backslash quad\; \backslash vartheta(x\_\{\backslash theta\},\; 1)\; =\; x\_0,\; \backslash end\{array\}\; \backslash right\backslash \}$$

for (1.8) is assumed. To any element $[\theta]$ belonging to the kernel of $\varphi$ let there correspond an element $[\xi_0]$ of the fundamental group $\pi_1(X)$ by the rule,

$$(2.2){\xi }_0(t)=\theta (x_0,t)\mathrm{.}(2.2)\; \backslash qquad\; \backslash xi\_0(t)\; =\; \backslash theta(x\_0,\; t).$$

This correspondence $\lambda$ has a definite meaning because, if $\theta \sim \theta'$, $\xi_0$ and $\xi_0'$ represent the same element of $\pi_1(X)$. Let us prove that $\lambda$ is an isomorphism. Let $[\theta_1], [\theta_2]$ be two elements belonging to the kernel of $\varphi$, then $[\theta_1] \times [\theta_2]$ is represented by a map $\theta_1 \times \theta_2$,

Since from (2.1) we have $\theta_2(x, 0) = x$, $\theta_2(\theta_1(x, 2t), 0) = \theta_1(x, 2t)$ and $\sigma_{\theta_2}(\theta_2(\theta_1(x, 1), 0), 2t-1) = \sigma_{\theta_2}(\theta_1(x, 1), 2t-1)$ so that by (2.2)

Since $\theta_1(x_0, 1) = x_0$ and $\sigma_{\theta_2}(x_0, t) = \theta_1(x_0, t)$, we have $\sigma_{\theta_2}(\theta_1(x_0, 1), 2t-1) = \theta_2(x_0, 2t-1)$. Now $\xi_{\theta_1 \times \theta_2}(t)$ may be described as follows:

On the other hand, we have, by the definition of the fundamental group,

$$\lambda ([{\theta }_1]\times [{\theta }_2])=[{\widehat{\xi }}_{{\theta }_1\times {\theta }_2}]=[{\widehat{\xi }}_{{\theta }_1}]\circ [{\widehat{\xi }}_{{\theta }_2}]=\lambda [{\theta }_1]\circ \lambda [{\theta }_2],\backslash lambda([\backslash theta\_1]\; \backslash times\; [\backslash theta\_2])\; =\; [\backslash hat\{\backslash xi\}\_\{\backslash theta\_1\; \backslash times\; \backslash theta\_2\}]\; =\; [\backslash hat\{\backslash xi\}\_\{\backslash theta\_1\}]\; \backslash circ\; [\backslash hat\{\backslash xi\}\_\{\backslash theta\_2\}]\; =\; \backslash lambda[\backslash theta\_1]\; \backslash circ\; \backslash lambda[\backslash theta\_2],$$

so that the homomorphism is established.

Clearly $\lambda$ is an onto-homomorphism, because of the homotopy extension property (1.3) i). It remains only to prove that from $\xi_{\theta_1} \sim \xi_{\theta_2}$ follows $\theta_1 \sim \theta_2$. It may be assumed that $\theta_1(x, 0) = x$ and $\theta_2(x, 0) = 0$. Since $\xi_{\theta_1} = \xi_{\theta_2}$, a homotopy $h_s(t)$ ($1 \le s \le 0$) exists such that $h_0(t) = \theta_1(x_0, t)$, $h_1(t) = \theta_2(x_0, t)$ and $h_s(0) = h_s(1) = x_0$. A continuous function $h$ may be defined on the set ${(X \times I)^s (0)^\tau [(X \times 0)^\tau (X \times 1)^\tau (x_0 \times I)] \times I}$ as follows:

$$h(x,s,0)=x,x\in X,s\in I^s,h(x,0,t)={\theta }_1(x,t),x\in X,t\in I^t,h(x,1,t)={\theta }_2(x,t),x\in X,t\in I^t,h(x_0,s,t)=h_s(t),s\in I^s,t\in I^t\mathrm{.}h(x,\; s,\; 0)\; =\; x,\; \backslash quad\; x\; \backslash in\; X,\; s\; \backslash in\; I^s,\; \backslash \backslash \; h(x,\; 0,\; t)\; =\; \backslash theta\_1(x,\; t),\; \backslash quad\; x\; \backslash in\; X,\; t\; \backslash in\; I^t,\; \backslash \backslash \; h(x,\; 1,\; t)\; =\; \backslash theta\_2(x,\; t),\; \backslash quad\; x\; \backslash in\; X,\; t\; \backslash in\; I^t,\; \backslash \backslash \; h(x\_0,\; s,\; t)\; =\; h\_s(t),\; \backslash quad\; s\; \backslash in\; I^s,\; t\; \backslash in\; I^t.$$

If (1.3) ii) is assumed, it is proved by the aid of the extended map $h: X \times I^s \times I^t$ ---PAGE_BREAK---

→ X that $\theta_1$ is homotopic to $\theta_2$. This completes the proof.

§ 3. Operation of $\mathfrak{A}(X)$ on the homotopy groups.

Let $f$ be a representative of an element $\alpha \in \pi_n(X)$ and $\theta$ be a representative of an element $\vartheta \in \mathfrak{A}(X)$. Let us define $\vartheta\alpha = [h] \in \pi_n(X)$ by the rule,

$$(3.1)h(x)\equiv \theta (f(x),1)\mathrm{.}(3.1)\; \backslash qquad\; h(x)\; \backslash equiv\; \backslash theta(f(x),\; 1).$$

This definition has a definite meaning in the sense that $[h]$ depends only on $\alpha$ and $\vartheta$. Then we have,

THEOREM 5. $\vartheta\alpha = (A\alpha)^{\xi}$ where $A = \varphi(\vartheta) \in \Sigma(X)$ and $\xi$ is an element of $\pi_1(X)$ represented by $\theta(x_0, t)$ ($1 \ge t \ge 0$).

Proof. From the definition of homomorphism $\varphi$, $A$ is represented by $a_0(x) = \theta(x, 0)$, and therefore $\theta(f(x), 0) = a_0f(x)$. It is an immediate consequence of the operation of $A$ that $a_0f$ represents an element $A\alpha$ of $\pi_n(X)$. Moreover if $f(p) = x_0$ for a fixed point $p \in S^n$, $\theta(f(p), t) = \theta(x_0, t)$ represents an element $\xi$ of $\pi_1(X)$, so that according to the operation of $\pi_1$ on $\pi_n$ due to Eilenberg $h(x) = \theta(f(x), 1)$ represents an element $(A\alpha)^{\xi} \in \pi_n$. This completes the proof.

As a direct consequence of Theorem 5 we have,

THEOREM 6. $\mathfrak{A}(X)$ is a group of automorphisms of $\pi_n(X)$ for every integer $n \ge 1$.

Proof. Because of the combination of automorphisms $A$ and $\xi$, the operation of $\vartheta \in \mathfrak{A}(X)$ on $\pi_n$ is also an automorphism of $\pi_n(X)$.

§ 4. Algebraic construction of $\mathfrak{A}(X)$.

Now that the operation of $\mathfrak{A}(X)$ on $\pi_n$ has been clarified by Theorem 5, we can construct the group $\mathfrak{A}(X)$ from a purely algebraic standpoint. Let $\chi(X) = {(A, \xi)}; A \in \Sigma(X), \xi \in \pi_1(X)}$; the totality of all the ordered pairs consisting of an arbitrarily chosen element of $\Sigma(X)$ and of an arbitrarily chosen element of $\pi_1(X)$. Defining $(A, \xi)(\alpha) = (A\alpha)^{\xi}$ for any $\alpha \in \pi_n(X)$, $(A, \xi)$ operates on $\pi_n(X)$, for every integer $\pi \ge 1$, as an automorphism. If we define a multiplication in the set $\chi(X)$ of automorphisms just defined by the rule,

$$(B,\eta )(A,\xi )(\alpha )=(B,\eta )((A,\xi )(\alpha )),(B,\; \backslash eta)(A,\; \backslash xi)(\backslash alpha)\; =\; (B,\; \backslash eta)((A,\; \backslash xi)(\backslash alpha)),$$

then we have $(B, \eta)(A, \xi) \in \chi(X)$. In order to prove this, we need the following lemma.

LEMMA 4.1 $A(\alpha^{\xi}) = (A\alpha)^{\xi} = (A, A_{\xi})(\alpha)$ for any $\alpha \in \pi_n$, where $A_{\xi}$ can be interpreted in the sense that $\Sigma(X) \ni A$ operates on the homotopy group of any dimension, especially on the fundamental group too.

Proof. Let $\alpha$ be represented by a mapping $f: S^n \to X, S^n \ni p_0 \to x_0$ and let ---PAGE_BREAK---

ξ = [e(t), 1 ≡ t ≡ 0]. We have a mapping F : {Sⁿ × (0) ⌣ (p₀) × I} → X such that F(x, 0) ≅ f(x) for any x ∈ Sⁿ, and F(p₀, t) ≅ e(t). From the homotopy extension property of a polyhedron we have an extended map $\bar{F}: S^n \times I \to X$ of F. Since $\bar{F}(x, 0) = f(x)$ and $\bar{F}(p_0, t) = e(t)$, $\bar{F}(x, 1)$ represents an element $a^t \in \pi_n(X)$. Let a be a representative of A. Putting $a(\bar{F}(x, t)) \equiv G(x, t): S^n \times I \to X$ we have $[G(x, 0)] = A\alpha$ from $G(x, 0) = a(f(x))$ and $[G(x, 1)] = A(\alpha^t)$ from $G(x, 1) = a(\bar{F}(x, 1))$. Also, from $G(x_0, t) = a(e(t))$ follows $[G(x_0, t)] = A\xi$. Thus we have $A(\alpha^t) = (A\alpha)^{A^t}$. Making use of the lemma, we have

Thus $(B, \eta)(A, \xi) = (A \cdot B, B\xi \cdot \eta) \in \chi(X).$

THEOREM 7. By this multiplication $\chi(X)$ forms a group.

Proof. As to the associative law we have

Thus $$(C,\zeta )((B,\eta )(A,\xi ))=((C,\zeta )(B,\eta ))(A,\xi )(C,\; \backslash zeta)((B,\; \backslash eta)(A,\; \backslash xi))\; =\; ((C,\; \backslash zeta)(B,\; \backslash eta))(A,\; \backslash xi)$$

The existence of the unity is proved as follows :

($E$, $e$)(A, $\xi$) = ($AE$, $E\xi \cdot e$) = (A, $\xi$) where E, e are the unities of $\Sigma(X)$ and $\pi_1(X)$ respectively.

The existence of an inverse element is proved thus :

$$(A^{-1},A^{-1}{\xi }^{-1})(A,\xi )=(A{A}^{-1},A^{-1}\xi \cdot A^{-1}{\xi }^{-1})=(E,A^{-1}(\xi {\xi }^{-1}))=(E,e)\mathrm{.}(A^\{-1\},\; A^\{-1\}\backslash xi^\{-1\})(A,\; \backslash xi)\; =\; (AA^\{-1\},\; A^\{-1\}\backslash xi\; \backslash cdot\; A^\{-1\}\backslash xi^\{-1\})\; =\; (E,\; A^\{-1\}(\backslash xi\backslash xi^\{-1\}))\; =\; (E,\; e).$$

This completes the proof.

Now the following MAIN THEOREM concerning the relation of two groups $\mathfrak{A}(X)$ and $\chi(X)$ imparts the complete analysis to the structure of $\mathfrak{A}(X)$ and also to the operation of $\mathfrak{A}(X)$ on $\pi_n(X)$ for every integer $n \ge 1$.

MAIN THEOREM 8. $\mathfrak{A}(x)$ is isomorphic to the group $\chi(X)$. Moreover, an isomorphism can be established between these groups, preserving the operation on the homotopy groups.

Proof. The method of proof being analogous as for Theorems 4, 5, we shall ---PAGE_BREAK---

restrict ourselves to show the correspondence between two groups. Let $\theta$ be a representative of $\partial \mathfrak{U}(X)$ and let $a_0 = \theta | X \times 0, \xi_0 = \theta | x_0 \times I$. Then to $\partial$ let there correspond $([a_0], [\xi_0]) \in \chi(X)$. It can be shown that this correspondence is an isomorphism and that the operations of $\partial$ and of the corresponding element $([a_0], [\xi_0])$ on $\pi_n$ are the same.

§ 5. Some remarks on the group $\mathfrak{U}(X)$.

By the aid of the main theorem it is advantageous to use $\chi(X)$ in place of $\mathfrak{U}(X)$ in calculating the invariant $\mathfrak{U}(X)$ of the space $X$. As is easily seen, two distinct elements of $\chi(X)$ do not always operate differently on $\pi_n$ so that as the group of the operation on $\pi_n$, $\chi(X)$ may be reduced to a smaller group. This reduction gives rise to an analogous classification of the space $X$ as the simplicity of a space due to Eilenberg.

Let $\chi^*(X)$ be the totality of all elements in $\chi(X)$ whose operations on any element of $\pi_n(X)$ are trivial; i.e. $\chi^*(X) = {(A, \xi) ; (A, \xi)(\alpha) = \alpha$ for any element $\alpha \in \pi_n(X)}$. Then $\chi^*(X)$ is clearly a normal subgroup of $\chi(X)$. Similarly, put $\chi^{}(X) = {(A, e) ; (A, e)(\alpha) = \alpha$ for any $\alpha \in \pi_n(X)}$ and $\chi^{}(X) = {(E, \xi) ; (E, \xi)(\alpha) = \alpha$ for any $\alpha \in \pi_n(X)}$, then these two groups are also normal in $\Sigma(X)$ and $\pi_1(X)$ respectively as well as in $\chi(X)$. It is well known that the space is $n$-simple in the sense of Eilenberg if $\chi^{}(X) \cong \pi_1(X)$. It may be an interesting problem to consider the spaces satisfying the conditions such as $\chi^(X) = \chi(X)$ or $\chi^{*}(X) \cong \Sigma(X)$.

BIBLIOGRAPHY

[1] Eilenberg, S., On the relation between the fundamental group of a space and higher homotopy groups, Fundamenta Math. 22 (1939).

[2] Hu, S. T., On the Whitehead Group of automorphisms of the relative homotopy groups, Portugaliae Math. 7 (1948).