[proofplan] The theorem is an "iff" with two independent implications that we prove in sequence. For the forward direction, given two paths $u, v$ with the same endpoints, we form the loop $u \cdot v^{-1}$ and use simple-connectedness to contract it to the constant loop; a short chain of standard path-homotopy identities (left/right cancellation, associativity, cancellation of inverses) then converts this contraction into a homotopy $u \simeq v$. For the reverse direction, we specialise the uniqueness of paths to the case $x_0 = x_1$: loops at $x_0$ are paths from $x_0$ to itself, and the constant loop $c_{x_0}$ is one such path, so uniqueness of the homotopy class forces every loop to be nullhomotopic, meaning $\pi_1(X, x_0) = \{e\}$. [/proofplan] [step:Fix notation for paths, reverses, concatenations, and based-path homotopy] We write a path in $X$ as a continuous map \begin{align*} u: I &\to X, & I := [0, 1], \end{align*} and denote $u: x_0 \rightsquigarrow x_1$ when $u(0) = x_0$ and $u(1) = x_1$. For a point $x \in X$, the constant path at $x$ is \begin{align*} c_x: I &\to X, & s &\mapsto x. \end{align*} For paths $u: x_0 \rightsquigarrow x_1$ and $v: x_1 \rightsquigarrow x_2$, their concatenation $u \cdot v$ is the path $x_0 \rightsquigarrow x_2$ defined by \begin{align*} (u \cdot v)(s) = \begin{cases} u(2s) & s \in [0, 1/2], \\ v(2s - 1) & s \in [1/2, 1]. \end{cases} \end{align*} The reverse of $u: x_0 \rightsquigarrow x_1$ is \begin{align*} u^{-1}: I &\to X, & s &\mapsto u(1 - s), \end{align*} which is a path $x_1 \rightsquigarrow x_0$. Throughout, $\simeq$ denotes path-homotopy (homotopy of paths with fixed endpoints). Recall from the [standard path-homotopy identities](/theorems/???): \begin{align*} u \cdot c_{x_1} &\simeq u, & c_{x_0} \cdot u &\simeq u, & u \cdot u^{-1} &\simeq c_{x_0}, & u^{-1} \cdot u &\simeq c_{x_1}, \end{align*} and path-homotopy respects concatenation: if $\alpha \simeq \alpha'$ and $\beta \simeq \beta'$ (with concatenations defined), then $\alpha \cdot \beta \simeq \alpha' \cdot \beta'$. Associativity up to path-homotopy also holds: $(\alpha \cdot \beta) \cdot \gamma \simeq \alpha \cdot (\beta \cdot \gamma)$ whenever the concatenations are defined. Recall further that $X$ is [simply connected](/page/Simply%20Connected) means $X$ is path-connected and $\pi_1(X, x_0) = \{e\}$ for some (equivalently every) basepoint $x_0$. [/step] [step:Prove the forward direction — from $\pi_1(X, x_0) = \{e\}$ to uniqueness of path-homotopy classes] Assume $X$ is simply connected and fix $x_0, x_1 \in X$. Since $X$ is path-connected, some path from $x_0$ to $x_1$ exists, so the set of path-homotopy classes from $x_0$ to $x_1$ is nonempty. We show it has at most one element. Let $u, v: x_0 \rightsquigarrow x_1$ be any two paths. The concatenation $u \cdot v^{-1}$ is a loop based at $x_0$ because $(u \cdot v^{-1})(0) = u(0) = x_0$ and $(u \cdot v^{-1})(1) = v^{-1}(1) = v(0) = x_0$. By hypothesis $\pi_1(X, x_0) = \{e\}$, so the class of every loop based at $x_0$ is the identity class, which is $[c_{x_0}]$. Thus \begin{align*} u \cdot v^{-1} \simeq c_{x_0}. \end{align*} Concatenate on the right by $v$, using that path-homotopy respects concatenation: \begin{align*} (u \cdot v^{-1}) \cdot v \simeq c_{x_0} \cdot v. \end{align*} On the left, apply associativity: $(u \cdot v^{-1}) \cdot v \simeq u \cdot (v^{-1} \cdot v)$. On the right, apply $c_{x_0} \cdot v \simeq v$. Chaining these, \begin{align*} u \cdot (v^{-1} \cdot v) \simeq v. \end{align*} Now $v^{-1} \cdot v \simeq c_{x_1}$ (cancellation of inverses at the endpoint), and concatenation respects path-homotopy, so $u \cdot (v^{-1} \cdot v) \simeq u \cdot c_{x_1}$. Combining with $u \cdot c_{x_1} \simeq u$ (right identity), \begin{align*} u \simeq u \cdot c_{x_1} \simeq u \cdot (v^{-1} \cdot v) \simeq v. \end{align*} Since $u, v$ were arbitrary paths from $x_0$ to $x_1$, all such paths are path-homotopic: there is at most one path-homotopy class, as claimed. [guided] Why does the "loop equals identity" hypothesis force two paths with common endpoints to be path-homotopic? The mechanism is cancellation: if $u \cdot v^{-1}$ is contractible, then concatenating $v$ on the right cancels the $v^{-1}$ (up to path-homotopy), leaving $u$ on one side and $v$ on the other. Each step of the chain $u \simeq u \cdot c_{x_1} \simeq u \cdot (v^{-1} \cdot v) \simeq (u \cdot v^{-1}) \cdot v \simeq c_{x_0} \cdot v \simeq v$ uses exactly one of the standard path-homotopy identities, so each $\simeq$ sign must be justified: - $u \simeq u \cdot c_{x_1}$: right identity axiom. - $u \cdot c_{x_1} \simeq u \cdot (v^{-1} \cdot v)$: respectful concatenation, since $c_{x_1} \simeq v^{-1} \cdot v$. - $u \cdot (v^{-1} \cdot v) \simeq (u \cdot v^{-1}) \cdot v$: associativity (applied backward). - $(u \cdot v^{-1}) \cdot v \simeq c_{x_0} \cdot v$: respectful concatenation, since $u \cdot v^{-1} \simeq c_{x_0}$ (the hypothesis). - $c_{x_0} \cdot v \simeq v$: left identity axiom. The chain takes $u$ to $v$ via five applications of standard identities. The one place the simple-connectedness hypothesis enters is the fourth step, where we replace the loop $u \cdot v^{-1}$ by the constant loop $c_{x_0}$. Everything else is formal manipulation that holds in any space. [/guided] [/step] [step:Prove the reverse direction — from uniqueness of path-homotopy classes to $\pi_1(X, x_0) = \{e\}$] Assume that $X$ is path-connected and that for every pair of points $y_0, y_1 \in X$ the set of path-homotopy classes of paths from $y_0$ to $y_1$ has exactly one element. (Path-connectedness guarantees that the set is nonempty; uniqueness is the additional hypothesis.) Fix any basepoint $x_0 \in X$. Specialise the hypothesis to the case $y_0 = y_1 = x_0$: there is a unique path-homotopy class of paths from $x_0$ to $x_0$, that is, a unique path-homotopy class of loops at $x_0$. The constant loop $c_{x_0}$ is a loop at $x_0$, so $[c_{x_0}]$ is one element of this set of classes. By uniqueness, every loop $\gamma: I \to X$ with $\gamma(0) = \gamma(1) = x_0$ satisfies $[\gamma] = [c_{x_0}]$. By definition, $\pi_1(X, x_0) = \{[\gamma] : \gamma \text{ loop at } x_0\}$, and the identity element of this group is $[c_{x_0}]$. We have just shown every element of $\pi_1(X, x_0)$ equals $[c_{x_0}]$, so $\pi_1(X, x_0) = \{[c_{x_0}]\} = \{e\}$. Combined with path-connectedness (which is part of the hypothesis), this says $X$ is simply connected. [guided] The reverse direction is the easy half. The uniqueness hypothesis is stated for all pairs $(y_0, y_1)$; specialising to $y_0 = y_1 = x_0$ is the only case we need. The constant loop $c_{x_0}$ serves as a witness that the unique class is the identity class. Without a witness we would only know that all loops are pairwise homotopic, not that they are homotopic to $c_{x_0}$ specifically — but $c_{x_0}$ is itself a loop, so it belongs to the unique class. The only non-pointed content of the hypothesis that we have used is that $\pi_1$ at $x_0$ is trivial; the statement "for every pair $(y_0, y_1)$" was stronger than we needed. Path-connectedness — which is also part of "simply connected" in our conventions — was already assumed. [/guided] [/step] [step:Conclude the equivalence] Step 2 shows "simply connected" implies "unique path-homotopy class of paths between any two points", and Step 3 shows the converse. Since $X$ is path-connected throughout (a stated hypothesis of the theorem, and a consequence of the "unique path-homotopy class" assumption which forces existence of at least one path), the two conditions are equivalent, completing the proof. [/step]