**Proof.** This result is usually stated in terms of cohomology, and for single spaces rather than pairs (for instance, Theorem VIII.2, §VIII.6, p. 149 in [7]). It can be derived from Theorem C.1 as follows. Let $g : X \cup C(A) \to S_n$ be defined as $g = f$ on $X$ and $g = s$ on $C(A)$. A null-homotopy of $f$ is a function $G : C(X) \to S_n$ extending $g$. The pair $(C(X), X \cup C(A))$ satisfies all the conditions of Hopf's extension theorem, so $g$ has extension if and only if $\ker i_* \subseteq \ker g_*$. As $C(X)$ is contractible, its reduced homology groups are trivial so $\ker i_*$ is the whole group $\tilde{H}_n(X \cup C(A); \mathbb{T})$; therefore, $g$ has an extension iff $g_* : \tilde{H}_n(X \cup C(A); \mathbb{T}) \to \tilde{H}_n(S_n; \mathbb{T})$ is trivial. Up to natural isomorphisms, $f_* = g_*$. $\square$

## D Coefficients

In Definition 4.1 we used the coefficient group $\mathbb{T}$ to detect relative cycles. One could equivalently use the groups $\mathbb{Z}_k = \mathbb{Z}/k\mathbb{Z}$ (with $k \ge 2$). The group $\mathbb{Z}$ can also give information. Although the next result is probably folklore, we did not find a suitable reference and include a proof for completeness.

**Proposition D.1.** Let $(X, A)$ be a compact pair and $x \in X \setminus A$ be $n$-Euclidean. The following statements are equivalent:

* The homomorphism $H_n(X, A; \mathbb{T}) \to H_n(X, X \setminus \{x\}; \mathbb{T})$ is non-trivial,

* The homomorphism $H_n(X, A; \mathbb{Z}_k) \to H_n(X, X \setminus \{x\}; \mathbb{Z}_k)$ is non-trivial for some $k \ge 2$,

and they hold if the homomorphism $H_n(X, A) \to H_n(X, X \setminus \{x\})$ is non-trivial.

**Proof.** We use the universal coefficient theorem for homology, and the fact that it is functional in the space and in the coefficient group. For any abelian group $G$, one has the following commuting diagram:

$$ \begin{tikzcd} 0 \arrow[r] & H_n(X, A) \otimes G \arrow[r, "g_1"] \arrow[d, "f_1"'] & H_n(X, A; G) \arrow[r, "g_2"] \arrow[d, "f_2"'] & \operatorname{Tor}(H_{n-1}(X, A), G) \arrow[r, "f_3"'] \arrow[d, "f_3"'] & 0 \\ 0 \arrow[r] & \underbrace{H_n(\mathbb{S}_n, s) \otimes G}_{G} \arrow[r, "h_1"] & \underbrace{H_n(\mathbb{S}_n, s; G)}_{G} \arrow[r, "h_2"] & \underbrace{\operatorname{Tor}(H_{n-1}(\mathbb{S}_n, s), G)}_{0} \arrow[r, "0"'] & 0 \end{tikzcd} \quad (10) $$

We will write $f_1^k$ for $G = \mathbb{Z}_k$ and $f_1'$ for $G = \mathbb{T}$, and similarly for the other homomorphisms.

We first assume that $f : H_n(X, A) \to H_n(\mathbb{S}_n, s) \cong \mathbb{Z}$ is non-trivial and show that $f_2^k$ is non-trivial for some $k \ge 2$. Let $q \in H_n(\mathbb{S}_n, s)$ be a non-zero element of the image of $f$. Let $k \ge 2$ be an integer that does not divide $q$, and take $G = \mathbb{Z}_k$ in (10), so that $(q, 1)$ is not zero in $H_n(\mathbb{S}_n, s) \otimes G$. As $(q, 1)$ belongs to the image of $f_1^k$, $f_1^k$ is non-trivial. As $h_1^k$ is injective, $h_1^k \circ f_1^k = f_2^k \circ g_1^k$ is non-trivial, so $f_2^k$ must be non-trivial.
be found in Hurewicz-Wallman [8]. They are stated in terms of Čech homology for compact spaces in [8]. We state them for compact ANRs using singular homology, which is equivalent to Čech homology for these spaces.

Hopf's extension theorem is Theorem VIII.1' in [8] (§VIII.6, p. 147).

**Theorem C.1** (Hopf's extension theorem). Let $n \in \mathbb{N}$. Let $(X, A)$ be a compact pair, where $X$ and $A$ are ANRs and $\dim(X) \le n + 1$. For a continuous function $f: A \to \mathbb{S}_n$, the following statements are equivalent:

* $f$ has a continuous extension $F: X \to \mathbb{S}_n$,

* $\ker i_* \subseteq \ker f_*$,

where $i: A \to X$ is the inclusion map and

$$i_*: H_n(A; \mathbb{T}) \to H_n(X; \mathbb{T})$$

$$f_*: H_n(A; \mathbb{T}) \to H_n(\mathbb{S}_n; \mathbb{T})$$

are the homomorphisms induced by $i$ and $f$.

The equivalence holds for $n = 0$ or $1$, without any dimension assumption about $X$.

For $n = 0$, Theorem C.1 also holds when considering homomorphisms between reduced homology groups $\tilde{H}_0(\cdot; \mathbb{T})$.

We present the argument for the particular case $n = 0$ or $1$, which is not stated in [8] but follows from the proof.

*Proof.* Assume that $n \le 1$. We only need to prove the result for CW-pairs $(X, A)$, because a pair of compact ANRs is homotopy equivalent to a CW-pair. We assume that $\ker i_* \subseteq \ker f_*$ and prove that $f$ has an extension. From Hopf's extension theorem for spaces of dimension at most $n+1$, $f$ has an extension to the $n+1$-skeleton of $X$. For $k \ge n+1$, every function from $\mathbb{S}_k$ to $\mathbb{S}_n$ is null-homotopic (because $n = 0$ or $1$), so $f$ can be inductively extended to every cell of $X$. $\square$

**Theorem C.2** (Hopf's classification theorem). Let $n \in \mathbb{N}$. Let $(X, A)$ be a compact pair, where $X$ and $A$ are ANRs and $\dim(X) \le n$. For a continuous function $f: (X, A) \to (\mathbb{S}_n, s)$, the following statements are equivalent:

* $f$ is null-homotopic (as a function of pairs),

* $f_*: \tilde{H}_n(X, A; \mathbb{T}) \to \tilde{H}_n(\mathbb{S}_n, s; \mathbb{T})$ is trivial.

The equivalence holds for $n \le 1$, without any dimension assumption about $X$.

When $n = 0$ and $A = \emptyset$, and only in that case, one indeed needs to consider the homomorphism between reduced homology groups.
We now assume that $f_2^k$ is non-trivial for some $k \ge 2$ and show that $f_2'$ is non-trivial. The inclusion $\mathbb{Z}_k \to \mathbb{T}$ sending $[p]$ to $[p/k]$ induces homomorphisms from $H_n(Y, B; \mathbb{Z}_k)$ to $H_n(Y, B; \mathbb{T})$ which make the following diagram commute:

$$ \begin{array}{ccc} H_n(X, A; \mathbb{Z}_k) & \xrightarrow{i_2} & H_n(X, A; \mathbb{T}) \\ \Big\downarrow_{f_2^k} & & \Big\downarrow_{f_2'} \\ H_n(\mathbb{S}_n, s; \mathbb{Z}_k) & \xrightarrow{j_2} & H_n(\mathbb{S}_n, s; \mathbb{T}) \end{array} \qquad (11) $$

Note that $j_2 : \mathbb{Z}_k \to \mathbb{T}$ is the inclusion so it is injective. Therefore, the non-triviality of $f_2^k$ implies the non-triviality of $f_2'$.

Finally, we assume that $f_2'$ is non-trivial and show that $f_2^k$ is non-trivial for some $k \ge 2$. Again, the inclusion $\mathbb{Z}_k \to \mathbb{T}$ sending $[m]$ to $[m/k]$ induces the following commuting diagram:

$$ \begin{array}{ccc} 0 & \longrightarrow & H_n(X, A) \otimes \mathbb{Z}_k \xrightarrow{g_1^k} H_n(X, A; \mathbb{Z}_k) \xrightarrow{g_2^k} \operatorname{Tor}(H_{n-1}(X, A), \mathbb{Z}_k) \longrightarrow 0 \\ \Big\downarrow_{i_1} & & \Big\downarrow_{i_2} \\ 0 & \longrightarrow & H_n(X, A) \otimes \mathbb{T} \xrightarrow{g_1'} H_n(X, A; \mathbb{T}) \xrightarrow{g_2'} \operatorname{Tor}(H_{n-1}(X, A), \mathbb{T}) \longrightarrow 0 \end{array} \qquad (12) $$

*Claim D.1.* For every $c \in H_n(X, A; \mathbb{T})$, there exist $k \ge 2$, $d \in H_n(X, A) \otimes \mathbb{T}$ and $e \in H_n(X, A; \mathbb{Z}_k)$ such that $c = g_1'(d) + i_2(e)$.

*Proof.* Note that $\operatorname{Tor}(H_{n-1}(X, A), \mathbb{T})$ is the torsion subgroup of $H_{n-1}(X, A)$. Therefore, $g_2'(c)$ is a torsion element of $H_{n-1}(X, A)$, so there exists $k \ge 2$ such that $g_2'(c) \in \operatorname{im}(i_3)$. As $g_2^k$ is surjective, $g_2'(c) = i_3 \circ g_2^k(e)$ for some $e \in H_n(X, A; \mathbb{Z}_k)$. One has $g_2' \circ i_2(e) = i_3 \circ g_2^k(e) = g_2'(c)$, so $c - i_2(e) \in \ker(g_2') = \operatorname{im}(g_1')$, therefore there exists $d \in H_n(X, A) \otimes \mathbb{T}$ such that $c - i_2(e) = g_1'(d)$. $\square$

Now assume that $f_2'$ is non-trivial, and let $c \in H_n(X, A; \mathbb{T})$ have a non-zero image under $f_2'$. One has $c = g_1'(d) + i_2(e)$ as in Claim D.1, so $f_2' \circ g_1'(d) \ne 0$ or $f_2' \circ i_2(e) \ne 0$. If $f_2' \circ g_1'(d) \ne 0$, then $f_1'$ is non-trivial by (10) so $f$ is non-trivial and we can apply the first argument of the proof, implying that $f_2^r$ is non-trivial for some $r \ge 2$. If $f_2' \circ i_2(e) \ne 0$, then $f_2^k$ is non-trivial by (11). $\square$

It can happen that the homomorphism is trivial with the coefficient group $\mathbb{Z}$ but not with some $\mathbb{Z}_k$. It happens for instance if $X$ is an $n$-manifold that is not orientable, with $k = 2$.
**Proof.** This result is usually stated in terms of cohomology, and for single spaces rather than pairs (for instance, Theorem VIII.2, §VIII.6, p. 149 in [8]). It can be derived from Theorem C.1 as follows. Let $g: X \cup C(A) \to S_n$ be defined as $g=f$ on $X$ and $g=s$ on $C(A)$. A null-homotopy of $f$ is a function $G: C(X) \to S_n$ extending $g$. The pair $(C(X), X \cup C(A))$ satisfies all the conditions of Hopf's extension theorem, so $g$ has extension if and only if $\ker i_* \subseteq \ker g_*$. As $C(X)$ is contractible, its reduced homology groups are trivial so $\ker i_*$ is the whole group $\tilde{H}_n(X \cup C(A); \mathbb{T})$; therefore, $g$ has an extension iff $g_* : \tilde{H}_n(X \cup C(A); \mathbb{T}) \to \tilde{H}_n(S_n; \mathbb{T})$ is trivial. Up to natural isomorphisms, $f_*$ is $g_*$. $\square$

# D Coefficients

In Definition 4.1 we used the coefficient group $\mathbb{T}$ to detect relative cycles. One could equivalently use the groups $\mathbb{Z}_k = \mathbb{Z}/k\mathbb{Z}$ (with $k \ge 2$). The group $\mathbb{Z}$ can also give information. Although the next result is probably folklore, we did not find a suitable reference and include a proof for completeness.

**Proposition D.1.** Let $(X, A)$ be a compact pair and $x \in X \setminus A$ be $n$-Euclidean. The following statements are equivalent:

* The homomorphism $H_n(X, A; \mathbb{T}) \to H_n(X, X \setminus \{x\}; \mathbb{T})$ is non-trivial,

* The homomorphism $H_n(X, A; \mathbb{Z}_k) \to H_n(X, X \setminus \{x\}; \mathbb{Z}_k)$ is non-trivial for some $k \ge 2$,

and they hold if the homomorphism $H_n(X, A) \to H_n(X, X \setminus \{x\})$ is non-trivial.

**Proof.** We use the universal coefficient theorem for homology, and the fact that it is functional in the space and in the coefficient group. For any abelian group $G$, one has the following commuting diagram:

$$ \begin{tikzcd} 0 \arrow[r] & H_n(X, A) \otimes G \arrow[r, "g_1"] \arrow[d, "f_1"'] & H_n(X, A; G) \arrow[r, "g_2"] \arrow[d, "f_2"'] & \operatorname{Tor}(H_{n-1}(X, A), G) \arrow[r, "f_3"'] \arrow[d, "f_3"'] & 0 \\ 0 \arrow[r] & \underbrace{H_n(\mathbb{S}_n, s) \otimes G}_{G} \arrow[r, "h_1"] & \underbrace{H_n(\mathbb{S}_n, s; G)}_{G} \arrow[r, "h_2"] & \underbrace{\operatorname{Tor}(H_{n-1}(\mathbb{S}_n, s), G)}_{0} \arrow[r, "0"'] & 0 \end{tikzcd} \quad (10) $$

We will write $f_1^k$ for $G = \mathbb{Z}_k$ and $f_1'$ for $G = \mathbb{T}$, and similarly for the other homomorphisms.

We first assume that $f: H_n(X, A) \to H_n(\mathbb{S}_n, s) \cong \mathbb{Z}$ is non-trivial and show that $f_2^k$ is non-trivial for some $k \ge 2$. Let $q \in H_n(\mathbb{S}_n, s)$ be a non-zero element of the image of $f$. Let $k \ge 2$ be an integer that does not divide $q$, and take $G = \mathbb{Z}_k$ in (10), so that $(q, 1)$ is not zero in $H_n(\mathbb{S}_n, s) \otimes G$. As $(q, 1)$ belongs to the image of $f_1^k$, $f_1^k$ is non-trivial. As $h_1^k$ is injective, $h_1^k \circ f_1^k = f_2^k \circ g_1^k$ is non-trivial, so $f_2^k$ must be non-trivial.
We now assume that $f_2^k$ is non-trivial for some $k \ge 2$ and show that $f_2'$ is non-trivial. The inclusion $\mathbb{Z}_k \to \mathbb{T}$ sending $[p]$ to $[p/k]$ induces homomorphisms from $H_n(Y, B; \mathbb{Z}_k)$ to $H_n(Y, B; \mathbb{T})$ which make the following diagram commute:

$$ \begin{array}{ccc} H_n(X, A; \mathbb{Z}_k) & \xrightarrow{i_2} & H_n(X, A; \mathbb{T}) \\ \Big\downarrow_{f_2^k} & & \Big\downarrow_{f_2'} \\ H_n(\mathbb{S}_n, s; \mathbb{Z}_k) & \xrightarrow{j_2} & H_n(\mathbb{S}_n, s; \mathbb{T}) \end{array} \qquad (11) $$

Note that $j_2 : \mathbb{Z}_k \to \mathbb{T}$ is the inclusion so it is injective. Therefore, the non-triviality of $f_2^k$ implies the non-triviality of $f_2'$.

Finally, we assume that $f_2'$ is non-trivial and show that $f_2^k$ is non-trivial for some $k \ge 2$. Again, the inclusion $\mathbb{Z}_k \to \mathbb{T}$ sending $[m]$ to $[m/k]$ induces the following commuting diagram:

$$ \begin{array}{ccc} 0 & \longrightarrow & H_n(X, A) \otimes \mathbb{Z}_k \xrightarrow{g_1^k} H_n(X, A; \mathbb{Z}_k) \xrightarrow{g_2^k} \operatorname{Tor}(H_{n-1}(X, A), \mathbb{Z}_k) \longrightarrow 0 \\ \Big\downarrow_{i_1} & & \Big\downarrow_{i_2} \\ 0 & \longrightarrow & H_n(X, A) \otimes \mathbb{T} \xrightarrow{g_1'} H_n(X, A; \mathbb{T}) \xrightarrow{g_2'} \operatorname{Tor}(H_{n-1}(X, A), \mathbb{T}) \longrightarrow 0 \end{array} \qquad (12) $$

**Claim D.1.** For every $c \in H_n(X, A; \mathbb{T})$, there exist $k \ge 2$, $d \in H_n(X, A) \otimes \mathbb{T}$ and $e \in H_n(X, A; \mathbb{Z}_k)$ such that $c = g_1'(d) + i_2(e)$.

**Proof.** Note that $\operatorname{Tor}(H_{n-1}(X, A), \mathbb{T})$ is the torsion subgroup of $H_{n-1}(X, A)$. Therefore, $g_2'(c)$ is a torsion element of $H_{n-1}(X, A)$, so there exists $k \ge 2$ such that $g_2'(c) \in \operatorname{im}(i_3)$. As $g_2^k$ is surjective, $g_2'(c) = i_3 \circ g_2^k(e)$ for some $e \in H_n(X, A; \mathbb{Z}_k)$. One has $g_2' \circ i_2(e) = i_3 \circ g_2^k(e) = g_2'(c)$, so $c - i_2(e) \in \ker(g_2') = \operatorname{im}(g_1')$, therefore there exists $d \in H_n(X, A) \otimes \mathbb{T}$ such that $c - i_2(e) = g_1'(d)$. $\square$

Now assume that $f_2'$ is non-trivial, and let $c \in H_n(X, A; \mathbb{T})$ have a non-zero image under $f_2'$. One has $c = g_1'(d) + i_2(e)$ as in Claim D.1, so $f_2' \circ g_1'(d) \ne 0$ or $f_2' \circ i_2(e) \ne 0$. If $f_2' \circ g_1'(d) \ne 0$, then $f_1'$ is non-trivial by (10) so $f$ is non-trivial and we can apply the first argument of the proof, implying that $f_2^r$ is non-trivial for some $r \ge 2$. If $f_2' \circ i_2(e) \ne 0$, then $f_2^k$ is non-trivial by (11). $\square$

It can happen that the homomorphism is trivial with the coefficient group $\mathbb{Z}$ but not with some $\mathbb{Z}_k$. It happens for instance if $X$ is an $n$-manifold that is not orientable, with $k = 2$.