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I’d argue that it boils down to the generator $S^n\to K(\mathbb{Z},n)$ being an $n$-equivalence.

More detail: If you take the represented version of homology, it is given by $$ H_n(X;\mathbb{Z}) \cong [ S^{n+t} , X\wedge K(\mathbb{Z}, t)] $$ for $t$ large. Then the Hurewicz map is the map induced by the generator $g:S^t \to K(\mathbb{Z}, t)$: $$ \mathrm{H}: [ S^{n+t} , X\wedge S^k] \xrightarrow{(\mathrm{id}_X\wedge g)_*} [ S^{n+t} , X\wedge K(\mathbb{Z}, t)]. $$ If $X$ is $(n-1)$-connected then since $g$ is a $t$-equivalence, the map $(\mathrm{id}_X\wedge g)_*$ is an isomorphism.

Even MORE detail: It is true that the existence of $K(\mathbb{Z},n)$ requires genuine work. But it does not require the Hurewicz theorem. In fact, you only need to produce an $(n-1)$-connected space $X$ with $\pi_n(X) \cong \mathbb{Z}$ (then you can take a Postnikov section to get $K(\mathbb{Z},n)$). And there is an obvious candidate: $S^n$.

So the question becomes, how can you show $\pi_n(S^n) \cong \mathbb{Z}$ for all $n$ (the connectivity is easy as a result of $S^n = * \cup D^n$ and the cellular approximation theorem). We can prove it for $n = 1$ using covering spaces, and use Freudenthal to get it for higher $n$ (it's a little tricky for $n= 2$, but luckily $S^1$ is an H-space).

What about proving Freudenthal? The suspension $\Sigma: \pi_n(X) \to \pi_{n+1}(\Sigma X)$ is equivalent to the map induced by $\sigma: X\to \Omega\Sigma X$, adjoint to $\mathrm{id}_{\Sigma X}$, and its connectivity can be estimated using the James construction.

What about the James construction? This lovely paper

Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389.

gives a proof based only on Mather's second cube theorem, which itself depends on some point-set topology due to Str{\o}m (no relation).

I’d argue that it boils down to the generator $S^n\to K(\mathbb{Z},n)$ being an $(n+1)$-equivalence.

More detail: If you take the represented version of homology, it is given by $$ H_n(X;\mathbb{Z}) \cong [ S^{n+t} , X\wedge K(\mathbb{Z}, t)] $$ for $t$ large. Then the Hurewicz map is the map induced by the generator $g:S^t \to K(\mathbb{Z}, t)$: $$ \mathrm{H}: [ S^{n+t} , X\wedge S^k] \xrightarrow{(\mathrm{id}_X\wedge g)_*} [ S^{n+t} , X\wedge K(\mathbb{Z}, t)]. $$ If $X$ is $(n-1)$-connected then since $g$ is a $(t+1)$-equivalence, the map $(\mathrm{id}_X\wedge g)_*$ is an isomorphism.

Even MORE detail: It is true that the existence of $K(\mathbb{Z},n)$ requires genuine work. But it does not require the Hurewicz theorem. In fact, you only need to produce an $(n-1)$-connected space $X$ with $\pi_n(X) \cong \mathbb{Z}$ (then you can take a Postnikov section to get $K(\mathbb{Z},n)$). And there is an obvious candidate: $S^n$.

So the question becomes, how can you show $\pi_n(S^n) \cong \mathbb{Z}$ for all $n$ (the connectivity is easy as a result of $S^n = * \cup D^n$ and the cellular approximation theorem). We can prove it for $n = 1$ using covering spaces, and use Freudenthal to get it for higher $n$ (it's a little tricky for $n= 2$, but luckily $S^1$ is an H-space).

What about proving Freudenthal? The suspension $\Sigma: \pi_n(X) \to \pi_{n+1}(\Sigma X)$ is equivalent to the map induced by $\sigma: X\to \Omega\Sigma X$, adjoint to $\mathrm{id}_{\Sigma X}$, and its connectivity can be estimated using the James construction.

What about the James construction? This lovely paper

Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389.

gives a proof based only on Mather's second cube theorem, which itself depends on some point-set topology due to Str{\o}m (no relation).

I’d argue that it boils down to the generator $S^n\to K(\mathbb{Z},n)$ being an $n$-equivalence.

More detail: If you take the represented version of homology, it is given by $$ H_n(X;\mathbb{Z}) \cong [ S^{n+t} , X\wedge K(\mathbb{Z}, t)] $$ for $t$ large. Then the Hurewicz map is the map induced by the generator $g:S^t \to K(\mathbb{Z}, t)$: $$ \mathrm{H}: [ S^{n+t} , X\wedge S^k] \xrightarrow{(\mathrm{id}_X\wedge g)_*} [ S^{n+t} , X\wedge K(\mathbb{Z}, t)]. $$ If $X$ is $(n-1)$-connected then since $g$ is a $(t+1)$-equivalence, the map $(\mathrm{id}_X\wedge g)_*$ is an isomorphism.

Even MORE detail: It is true that the existence of $K(\mathbb{Z},n)$ requires genuine work. But it does not require the Hurewicz theorem. In fact, you only need to produce an $(n-1)$-connected space $X$ with $\pi_n(X) \cong \mathbb{Z}$ (then you can take a Postnikov section to get $K(\mathbb{Z},n)$). And there is an obvious candidate: $S^n$.

So the question becomes, how can you show $\pi_n(S^n) \cong \mathbb{Z}$ for all $n$ (the connectivity is easy as a result of $S^n = * \cup D^n$ and the cellular approximation theorem). We can prove it for $n = 1$ using covering spaces, and use Freudenthal to get it for higher $n$ (it's a little tricky for $n= 2$, but luckily $S^1$ is an H-space).

What about proving Freudenthal? The suspension $\Sigma: \pi_n(X) \to \pi_{n+1}(\Sigma X)$ is equivalent to the map induced by $\sigma: X\to \Omega\Sigma X$, adjoint to $\mathrm{id}_{\Sigma X}$, and its connectivity can be estimated using the James construction.

What about the James construction? This lovely paper

Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389.

gives a proof based only on Mather's second cube theorem, which itself depends on some point-set topology due to Str{\o}m (no relation).

I’d argue that it boils down to the generator $S^n\to K(\mathbb{Z},n)$ being an $n$-equivalence.

More detail: If you take the represented version of homology, it is given by $$ H_n(X;\mathbb{Z}) \cong [ S^{n+t} , X\wedge K(\mathbb{Z}, t)] $$ for $t$ large. Then the Hurewicz map is the map induced by the generator $g:S^t \to K(\mathbb{Z}, t)$: $$ \mathrm{H}: [ S^{n+t} , X\wedge S^k] \xrightarrow{(\mathrm{id}_X\wedge g)_*} [ S^{n+t} , X\wedge K(\mathbb{Z}, t)]. $$ If $X$ is $(n-1)$-connected then since $g$ is a $t$-equivalence, the map $(\mathrm{id}_X\wedge g)_*$ is an isomorphism.

Even MORE detail: It is true that the existence of $K(\mathbb{Z},n)$ requires genuine work. But it does not require the Hurewicz theorem. In fact, you only need to produce an $(n-1)$-connected space $X$ with $\pi_n(X) \cong \mathbb{Z}$ (then you can take a Postnikov section to get $K(\mathbb{Z},n)$). And there is an obvious candidate: $S^n$.

So the question becomes, how can you show $\pi_n(S^n) \cong \mathbb{Z}$ for all $n$ (the connectivity is easy as a result of $S^n = * \cup D^n$ and the cellular approximation theorem). We can prove it for $n = 1$ using covering spaces, and use Freudenthal to get it for higher $n$ (it's a little tricky for $n= 2$, but luckily $S^1$ is an H-space).

What about proving Freudenthal? The suspension $\Sigma: \pi_n(X) \to \pi_{n+1}(\Sigma X)$ is equivalent to the map induced by $\sigma: X\to \Omega\Sigma X$, adjoint to $\mathrm{id}_{\Sigma X}$, and its connectivity can be estimated using the James construction.

What about the James construction? This lovely paper

Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389.

gives a proof based only on Mather's second cube theorem, which itself depends on some point-set topology due to Str{\o}m (no relation).

I’d argue that it boils down to the generator $S^n\to K(\mathbb{Z},n)$ being an $n$-equivalence.

More detail: If you take the represented version of homology, it is given by $$ H_n(X;\mathbb{Z}) \cong [ S^{n+t} , X\wedge K(\mathbb{Z}, t)] $$ for $t$ large. Then the Hurewicz map is the map induced by the generator $g:S^t \to K(\mathbb{Z}, t)$: $$ \mathrm{H}: [ S^{n+t} , X\wedge S^k] \xrightarrow{(\mathrm{id}_X\wedge g)_*} [ S^{n+t} , X\wedge K(\mathbb{Z}, t)]. $$ If $X$ is $(n-1)$-connected then since $g$ is a $(t+1)$-equivalence, the map $(\mathrm{id}_X\wedge g)_*$ is an isomorphism.

Even MORE detail: It is true that the existence of $K(\mathbb{Z},n)$ requires genuine work. But it does not require the Hurewicz theorem. In fact, you only need to produce an $(n-1)$-connected space $X$ with $\pi_n(X) \cong \mathbb{Z}$. And there is an obvious candidate: $S^n$.

So the question becomes, how can you show $\pi_n(S^n) \cong \mathbb{Z}$ for all $n$ (the connectivity is easy as a result of $S^n = * \cup D^n$ and the cellular approximation theorem). We can prove it for $n = 1$ using covering spaces, and use Freudenthal to get it for higher $n$ (it's a little tricky for $n= 2$, but luckily $S^1$ is an H-space).

What about proving Freudenthal? The suspension $\Sigma: \pi_n(X) \to \pi_{n+1}(\Sigma X)$ is equivalent to the map induced by $\sigma: X\to \Omega\Sigma X$, adjoint to $\mathrm{id}_{\Sigma X}$, and its connectivity can be estimated using the James construction.

What about the James construction? This lovely paper

Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389.

gives a proof based only on Mather's second cube theorem, which itself depends on some point-set topology due to Str{\o}m (no relation).

I’d argue that it boils down to the generator $S^n\to K(\mathbb{Z},n)$ being an $n$-equivalence.

More detail: If you take the represented version of homology, it is given by $$ H_n(X;\mathbb{Z}) \cong [ S^{n+t} , X\wedge K(\mathbb{Z}, t)] $$ for $t$ large. Then the Hurewicz map is the map induced by the generator $g:S^t \to K(\mathbb{Z}, t)$: $$ \mathrm{H}: [ S^{n+t} , X\wedge S^k] \xrightarrow{(\mathrm{id}_X\wedge g)_*} [ S^{n+t} , X\wedge K(\mathbb{Z}, t)]. $$ If $X$ is $(n-1)$-connected then since $g$ is a $(t+1)$-equivalence, the map $(\mathrm{id}_X\wedge g)_*$ is an isomorphism.

Even MORE detail: It is true that the existence of $K(\mathbb{Z},n)$ requires genuine work. But it does not require the Hurewicz theorem. In fact, you only need to produce an $(n-1)$-connected space $X$ with $\pi_n(X) \cong \mathbb{Z}$ (then you can take a Postnikov section to get $K(\mathbb{Z},n)$). And there is an obvious candidate: $S^n$.

So the question becomes, how can you show $\pi_n(S^n) \cong \mathbb{Z}$ for all $n$ (the connectivity is easy as a result of $S^n = * \cup D^n$ and the cellular approximation theorem). We can prove it for $n = 1$ using covering spaces, and use Freudenthal to get it for higher $n$ (it's a little tricky for $n= 2$, but luckily $S^1$ is an H-space).

What about proving Freudenthal? The suspension $\Sigma: \pi_n(X) \to \pi_{n+1}(\Sigma X)$ is equivalent to the map induced by $\sigma: X\to \Omega\Sigma X$, adjoint to $\mathrm{id}_{\Sigma X}$, and its connectivity can be estimated using the James construction.

What about the James construction? This lovely paper

Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389.

gives a proof based only on Mather's second cube theorem, which itself depends on some point-set topology due to Str{\o}m (no relation).