[proofplan] We first verify that $u_\#$ is well defined on homotopy classes using the pasting of path homotopies; the rel-$\{0, 1\}$ condition on $\gamma$ ensures the sandwich $u^{-1} \cdot \gamma \cdot u$ transports rigidly. Next we show $u_\#$ is a homomorphism: the key is that $u \cdot u^{-1} \simeq c_{x_1}$ allows two copies of $u \cdot u^{-1}$ inserted between $\gamma_0$ and $\gamma_1$ to cancel. The inverse homomorphism is $(u^{-1})_\#$, giving the isomorphism statement. The five numbered properties are then each verified by applying the [group laws for paths up to homotopy](/theorems/1877) to massage $u^{-1} \cdot \gamma \cdot u$ into the required form. Property 5, conjugation, is an immediate instance of 2 when $x_0 = x_1$. [/proofplan]

[step:Set up notation and record the basic moves] Throughout the proof, $I = [0, 1]$. The concatenation and reversal of paths are the standard operations. For a loop $\gamma$ at $x_0$, the triple concatenation is \begin{align*} (u^{-1} \cdot \gamma \cdot u)(s) = \begin{cases} u^{-1}(4s) = u(1 - 4s) & 0 \le s \le 1/4, \\ \gamma(4s - 1) & 1/4 \le s \le 1/2, \\ u(2s - 1) & 1/2 \le s \le 1, \end{cases} \end{align*} obtained by interpreting $u^{-1} \cdot \gamma \cdot u$ as $(u^{-1} \cdot \gamma) \cdot u$; by the associativity law of [Theorem 1877](/theorems/1877), this is path-homotopic to $u^{-1} \cdot (\gamma \cdot u)$, and we write it unparenthesised henceforth. We will repeatedly apply the [group laws for paths up to homotopy](/theorems/1877), abbreviated PHL: - **PHL-assoc**: $(\alpha \cdot \beta) \cdot \delta \simeq \alpha \cdot (\beta \cdot \delta)$ rel $\{0, 1\}$ - **PHL-id**: $\alpha \cdot c_y \simeq \alpha \simeq c_x \cdot \alpha$ rel $\{0, 1\}$ when $\alpha: x \rightsquigarrow y$ - **PHL-inv**: $\alpha \cdot \alpha^{-1} \simeq c_x$ and $\alpha^{-1} \cdot \alpha \simeq c_y$ rel $\{0, 1\}$ when $\alpha: x \rightsquigarrow y$ We will also use that [concatenation descends to classes](/theorems/1878) and is compatible with path homotopies — that is, if $\alpha \simeq \alpha'$ rel $\{0, 1\}$ and $\beta \simeq \beta'$ rel $\{0, 1\}$ are composable, then $\alpha \cdot \beta \simeq \alpha' \cdot \beta'$ rel $\{0, 1\}$. This is the "horizontal pasting" established in the well-definedness step of [Theorem 1878](/theorems/1878); it applies unchanged here, where the common endpoint is $x_1$ rather than the basepoint. [/step]

[step:Verify that $u_\#$ is well defined on path homotopy classes] Suppose $[\gamma] = [\gamma']$ in $\pi_1(X, x_0)$, witnessed by a path homotopy $H: I \times I \to X$ with $H(s, 0) = \gamma(s)$, $H(s, 1) = \gamma'(s)$, and $H(0, t) = H(1, t) = x_0$ for all $t \in I$. The path $u^{-1}$ is a path from $x_1$ to $x_0$, and $u$ a path from $x_0$ to $x_1$. Consider the constant homotopies $U^-: I \times I \to X$, $(s, t) \mapsto u^{-1}(s)$ and $U: I \times I \to X$, $(s, t) \mapsto u(s)$. These are path homotopies rel $\{0, 1\}$ from $u^{-1}$ to $u^{-1}$ and from $u$ to $u$ respectively. Horizontally concatenating $U^-$, $H$, $U$ (possible because the endpoint matchings $u^{-1}(1) = x_0 = \gamma(0)$ and $\gamma(1) = x_0 = u(0)$ hold uniformly in $t$ thanks to the rel-endpoints condition on $H$) yields a path homotopy \begin{align*} u^{-1} \cdot \gamma \cdot u \simeq u^{-1} \cdot \gamma' \cdot u \quad \text{rel } \{0, 1\}. \end{align*} Explicitly, let $K: I \times I \to X$ be defined by \begin{align*} K(s, t) = \begin{cases} u^{-1}(4s) & 0 \le s \le 1/4, \\ H(4s - 1, t) & 1/4 \le s \le 1/2, \\ u(2s - 1) & 1/2 \le s \le 1. \end{cases} \end{align*} The three closed pieces cover $I \times I$ and agree at their overlaps: at $s = 1/4$, the first gives $u^{-1}(1) = x_0$ and the second gives $H(0, t) = x_0$; at $s = 1/2$, the second gives $H(1, t) = x_0$ and the third gives $u(0) = x_0$. Each piece is continuous; the pasting lemma gives continuity of $K$. At $t = 0$, $K(\cdot, 0) = u^{-1} \cdot \gamma \cdot u$; at $t = 1$, $K(\cdot, 1) = u^{-1} \cdot \gamma' \cdot u$. At the $s$-endpoints, $K(0, t) = u^{-1}(0) = x_1$ and $K(1, t) = u(1) = x_1$, so $K$ is rel $\{0, 1\}$. Hence $[u^{-1} \cdot \gamma \cdot u] = [u^{-1} \cdot \gamma' \cdot u]$ in $\pi_1(X, x_1)$. The function \begin{align*} u_\#: \pi_1(X, x_0) \to \pi_1(X, x_1), \qquad [\gamma] \mapsto [u^{-1} \cdot \gamma \cdot u] \end{align*} is therefore well defined. [/step]

[step:Verify that $u_\#$ is a group homomorphism]Let $[\gamma_0], [\gamma_1] \in \pi_1(X, x_0)$. Then \begin{align*} u_\#([\gamma_0]) \cdot u_\#([\gamma_1]) = [u^{-1} \cdot \gamma_0 \cdot u] \cdot [u^{-1} \cdot \gamma_1 \cdot u] = [(u^{-1} \cdot \gamma_0 \cdot u) \cdot (u^{-1} \cdot \gamma_1 \cdot u)]. \end{align*} Apply PHL-assoc to rewrite the right-hand path (all steps are rel-$\{0, 1\}$ path homotopies in $X$): \begin{align*} (u^{-1} \cdot \gamma_0 \cdot u) \cdot (u^{-1} \cdot \gamma_1 \cdot u) &\simeq u^{-1} \cdot \gamma_0 \cdot (u \cdot u^{-1}) \cdot \gamma_1 \cdot u. \end{align*} By PHL-inv, $u \cdot u^{-1} \simeq c_{x_1}$. Using compatibility of concatenation with path homotopies, substitute $u \cdot u^{-1}$ by $c_{x_1}$: \begin{align*} u^{-1} \cdot \gamma_0 \cdot (u \cdot u^{-1}) \cdot \gamma_1 \cdot u \simeq u^{-1} \cdot \gamma_0 \cdot c_{x_1} \cdot \gamma_1 \cdot u. \end{align*} By PHL-id applied to $\gamma_0: x_0 \rightsquigarrow x_0$ (note $\gamma_0(1) = x_0 \ne x_1$ in general, so we cannot absorb $c_{x_1}$ into $\gamma_0$ directly — instead, we absorb it into $\gamma_1$ on the **left**, since $\gamma_1(0) = x_0$, not $x_1$). Let us be more careful: in the sandwich $u^{-1} \cdot \gamma_0 \cdot c_{x_1} \cdot \gamma_1 \cdot u$, the constant path $c_{x_1}$ is inserted between $\gamma_0: x_0 \to x_0$ and $\gamma_1: x_0 \to x_0$. But then $\gamma_0 \cdot c_{x_1}$ is not even defined as a concatenation, because $\gamma_0$ ends at $x_0$ and $c_{x_1}$ starts at $x_1$. We must re-examine the derivation. Re-deriving: the composable form requires the constant path between $\gamma_0$ and $\gamma_1$ to be at $x_0$, not $x_1$. We correct by inserting $u \cdot u^{-1}$ where the adjacent paths end/start at $x_1$. Read the sandwich: \begin{align*} (u^{-1} \cdot \gamma_0 \cdot u) \cdot (u^{-1} \cdot \gamma_1 \cdot u) \end{align*} is the concatenation of the loop $u^{-1} \cdot \gamma_0 \cdot u$ at $x_1$ with the loop $u^{-1} \cdot \gamma_1 \cdot u$ at $x_1$. By PHL-assoc we may regroup this loop at $x_1$ as \begin{align*} u^{-1} \cdot \bigl(\gamma_0 \cdot (u \cdot u^{-1}) \cdot \gamma_1\bigr) \cdot u, \end{align*} where the inner piece $\gamma_0 \cdot (u \cdot u^{-1}) \cdot \gamma_1$ is a valid concatenation: $\gamma_0$ ends at $x_0$, $u$ starts at $x_0$ (so $u \cdot u^{-1}$ starts at $x_0$), $u \cdot u^{-1}$ ends at $x_0$, and $\gamma_1$ starts at $x_0$. By PHL-inv, $u \cdot u^{-1} \simeq c_{x_0}$ rel $\{0, 1\}$. Substituting via compatibility, \begin{align*} \gamma_0 \cdot (u \cdot u^{-1}) \cdot \gamma_1 \simeq \gamma_0 \cdot c_{x_0} \cdot \gamma_1. \end{align*} By PHL-id applied to $\gamma_0: x_0 \rightsquigarrow x_0$, $\gamma_0 \cdot c_{x_0} \simeq \gamma_0$. Substituting and using PHL-assoc again, $\gamma_0 \cdot c_{x_0} \cdot \gamma_1 \simeq \gamma_0 \cdot \gamma_1$. Combining all these rel-$\{0, 1\}$ path homotopies and transporting through the outer $u^{-1}$ and $u$ by the compatibility of concatenation, \begin{align*} (u^{-1} \cdot \gamma_0 \cdot u) \cdot (u^{-1} \cdot \gamma_1 \cdot u) \simeq u^{-1} \cdot (\gamma_0 \cdot \gamma_1) \cdot u \quad \text{rel } \{0, 1\}. \end{align*} Taking classes, \begin{align*} u_\#([\gamma_0]) \cdot u_\#([\gamma_1]) = [u^{-1} \cdot (\gamma_0 \cdot \gamma_1) \cdot u] = u_\#([\gamma_0 \cdot \gamma_1]) = u_\#([\gamma_0] \cdot [\gamma_1]). \end{align*} So $u_\#$ is a homomorphism.[/step]

[guided]The homomorphism computation boils down to one idea: **conjugation by a path cancels on its own**. When we concatenate $u^{-1} \cdot \gamma_0 \cdot u$ with $u^{-1} \cdot \gamma_1 \cdot u$, the two adjacent copies $u$ and $u^{-1}$ in the middle annihilate up to homotopy, leaving $u^{-1} \cdot \gamma_0 \cdot \gamma_1 \cdot u$. **Setting up the computation.** Start with \begin{align*} u_\#([\gamma_0]) \cdot u_\#([\gamma_1]) = [u^{-1} \cdot \gamma_0 \cdot u] \cdot [u^{-1} \cdot \gamma_1 \cdot u]. \end{align*} Passing to a representative via well-definedness of concatenation on classes, \begin{align*} = [(u^{-1} \cdot \gamma_0 \cdot u) \cdot (u^{-1} \cdot \gamma_1 \cdot u)]. \end{align*} **Regrouping.** By PHL-assoc (applied repeatedly to compose the outer $u^{-1}$ and $u$), we may rewrite this as \begin{align*} [u^{-1} \cdot (\gamma_0 \cdot (u \cdot u^{-1}) \cdot \gamma_1) \cdot u]. \end{align*} **Checking composability.** The inner string $\gamma_0 \cdot (u \cdot u^{-1}) \cdot \gamma_1$ makes sense: $\gamma_0$ ends at $x_0$; $u \cdot u^{-1}$ starts at $x_0$ (since $u$ does), ends at $x_0$, and travels up to $x_1$ and back; $\gamma_1$ starts at $x_0$. Everything composes at $x_0$. Note the constant path we produce next is $c_{x_0}$, **not** $c_{x_1}$ — this was the point of caution in the exact version: $u \cdot u^{-1}$ is a loop at the **starting** point of $u$, which is $x_0$. **Applying PHL-inv.** PHL-inv gives $u \cdot u^{-1} \simeq c_{x_0}$ rel $\{0, 1\}$. By the compatibility of concatenation with path homotopies (established in the well-definedness step of [Theorem 1878](/theorems/1878)), \begin{align*} \gamma_0 \cdot (u \cdot u^{-1}) \cdot \gamma_1 \simeq \gamma_0 \cdot c_{x_0} \cdot \gamma_1 \quad \text{rel } \{0, 1\}. \end{align*} **Applying PHL-id.** PHL-id gives $\gamma_0 \cdot c_{x_0} \simeq \gamma_0$ rel $\{0, 1\}$, so $\gamma_0 \cdot c_{x_0} \cdot \gamma_1 \simeq \gamma_0 \cdot \gamma_1$ rel $\{0, 1\}$ by compatibility. **Transporting through the sandwich.** The outer $u^{-1}$ and $u$ are constant across the homotopy (no change there), and compatibility of concatenation applies: \begin{align*} u^{-1} \cdot (\gamma_0 \cdot \gamma_1) \cdot u \end{align*} is path-homotopic to the original expression. **Passing back to classes.** In $\pi_1(X, x_1)$: \begin{align*} u_\#([\gamma_0]) \cdot u_\#([\gamma_1]) = [u^{-1} \cdot (\gamma_0 \cdot \gamma_1) \cdot u] = u_\#([\gamma_0 \cdot \gamma_1]) = u_\#([\gamma_0] \cdot [\gamma_1]). \end{align*} This is the homomorphism property. **Why we are careful about the basepoint of the constant path.** In a first pass it is tempting to say "$u \cdot u^{-1} \simeq c_{x_1}$" because $u$ travels from $x_0$ to $x_1$, so one might expect a "loop" at $x_1$. But $u \cdot u^{-1}$ is the path that goes $x_0 \to x_1 \to x_0$, a loop based at $x_0$. PHL-inv gives $u \cdot u^{-1} \simeq c_{x_0}$ and separately $u^{-1} \cdot u \simeq c_{x_1}$. In the middle of our sandwich we have the former, not the latter.[/guided]

[step:Verify that $u_\#$ is bijective with inverse $(u^{-1})_\#$] By Steps 1–2 applied with the path $u^{-1}: x_1 \rightsquigarrow x_0$ in place of $u$, we obtain a well-defined group homomorphism \begin{align*} (u^{-1})_\#: \pi_1(X, x_1) \to \pi_1(X, x_0), \qquad [\delta] \mapsto [(u^{-1})^{-1} \cdot \delta \cdot u^{-1}] = [u \cdot \delta \cdot u^{-1}], \end{align*} where we used the pointwise identity $(u^{-1})^{-1}(s) = u^{-1}(1 - s) = u(s)$. Compute the composition: \begin{align*} ((u^{-1})_\# \circ u_\#)([\gamma]) = [u \cdot (u^{-1} \cdot \gamma \cdot u) \cdot u^{-1}]. \end{align*} By PHL-assoc, the path on the right is path-homotopic rel $\{0, 1\}$ to $(u \cdot u^{-1}) \cdot \gamma \cdot (u \cdot u^{-1})$. By PHL-inv, $u \cdot u^{-1} \simeq c_{x_0}$, so by compatibility \begin{align*} (u \cdot u^{-1}) \cdot \gamma \cdot (u \cdot u^{-1}) \simeq c_{x_0} \cdot \gamma \cdot c_{x_0} \simeq \gamma \quad \text{rel } \{0, 1\}, \end{align*} applying PHL-id twice. Therefore $((u^{-1})_\# \circ u_\#)([\gamma]) = [\gamma] = \operatorname{id}_{\pi_1(X, x_0)}([\gamma])$ for every $[\gamma]$. Symmetrically, replacing $u$ by $u^{-1}$ and $x_0 \leftrightarrow x_1$ in the same argument, $(u_\# \circ (u^{-1})_\#) = \operatorname{id}_{\pi_1(X, x_1)}$. Hence $u_\#$ is a bijection (homomorphism with two-sided inverse), and therefore a group isomorphism. [/step]

[step:Prove property (1): if $u \simeq u'$ then $u_\# = u'_\#$] Suppose $u \simeq u'$ rel $\{0, 1\}$, i.e., there exists a path homotopy from $u$ to $u'$. Reversing homotopy parameter, $u^{-1} \simeq (u')^{-1}$ rel $\{0, 1\}$: if $F(s, t)$ is a path homotopy from $u$ to $u'$, then $F(1 - s, t)$ is a path homotopy from $u^{-1}$ to $(u')^{-1}$, which is continuous and preserves endpoints ($F(1, t) = x_1 \to F(0, t) = x_0$ after reversal). Let $\gamma$ be a loop at $x_0$. By the compatibility of concatenation with path homotopies (applied three times, stacking the homotopies $u^{-1} \simeq (u')^{-1}$, the constant homotopy of $\gamma$, and $u \simeq u'$), \begin{align*} u^{-1} \cdot \gamma \cdot u \simeq (u')^{-1} \cdot \gamma \cdot u' \quad \text{rel } \{0, 1\}. \end{align*} Taking classes, $u_\#([\gamma]) = u'_\#([\gamma])$ for every class, so $u_\# = u'_\#$. [/step]

[step:Prove property (2): $(c_{x_0})_\# = \operatorname{id}_{\pi_1(X, x_0)}$] The path $c_{x_0}$ is a path from $x_0$ to $x_0$, and $c_{x_0}^{-1}(s) = c_{x_0}(1 - s) = x_0 = c_{x_0}(s)$, so $c_{x_0}^{-1} = c_{x_0}$ as maps. For a loop $\gamma$ at $x_0$, by PHL-id applied twice, \begin{align*} c_{x_0} \cdot \gamma \cdot c_{x_0} \simeq c_{x_0} \cdot \gamma \simeq \gamma \quad \text{rel } \{0, 1\}. \end{align*} Therefore $(c_{x_0})_\#([\gamma]) = [c_{x_0}^{-1} \cdot \gamma \cdot c_{x_0}] = [\gamma] = \operatorname{id}_{\pi_1(X, x_0)}([\gamma])$ for every class. [/step]

[step:Prove property (3): $(u \cdot v)_\# = v_\# \circ u_\#$]Let $v: x_1 \rightsquigarrow x_2$. Observe that $(u \cdot v)^{-1} \simeq v^{-1} \cdot u^{-1}$ rel $\{0, 1\}$: this follows because both paths begin at $x_2$, end at $x_0$, and have the pointwise equality \begin{align*} (u \cdot v)^{-1}(s) = (u \cdot v)(1 - s) = \begin{cases} v(1 - 2s) = v^{-1}(2s) & 0 \le s \le 1/2, \\ u(1 - (2s - 1)) = u^{-1}(2s - 1) & 1/2 \le s \le 1, \end{cases} \end{align*} which is exactly $(v^{-1} \cdot u^{-1})(s)$. So the two are **equal**, hence certainly homotopic. For a loop $\gamma$ at $x_0$, \begin{align*} (u \cdot v)_\#([\gamma]) = [(u \cdot v)^{-1} \cdot \gamma \cdot (u \cdot v)] = [v^{-1} \cdot u^{-1} \cdot \gamma \cdot u \cdot v], \end{align*} where the last equality uses the equality above and PHL-assoc to regroup. Now \begin{align*} [v^{-1} \cdot u^{-1} \cdot \gamma \cdot u \cdot v] = [v^{-1} \cdot (u^{-1} \cdot \gamma \cdot u) \cdot v] = v_\#([u^{-1} \cdot \gamma \cdot u]) = v_\#(u_\#([\gamma])) = (v_\# \circ u_\#)([\gamma]), \end{align*} where we used PHL-assoc to parenthesise as $v^{-1} \cdot (u^{-1} \cdot \gamma \cdot u) \cdot v$ (noting the inner piece is a loop at $x_1$, the starting point of $v$), the definition of $v_\#$, and the definition of $u_\#$. This holds for every class, so $(u \cdot v)_\# = v_\# \circ u_\#$.[/step]

[guided]We want to show that the change-of-basepoint map respects composition of paths, but with the **order reversed**: $(u \cdot v)_\# = v_\# \circ u_\#$ rather than $u_\# \circ v_\#$. This anti-homomorphism pattern happens because conjugation by a product reverses: $(uv)^{-1} x (uv) = v^{-1}(u^{-1} x u) v$, exactly the shape of $v_\# \circ u_\#$. The proof has two ingredients. **Ingredient 1.** The reverse of a concatenation is the concatenation of reverses in the opposite order: $(u \cdot v)^{-1} = v^{-1} \cdot u^{-1}$ **pointwise**. This is a direct computation with the piecewise concatenation formula (see the exact version). No homotopy needed — it is literal equality of functions $I \to X$. **Ingredient 2.** Substituting into the definition of $(u \cdot v)_\#$: \begin{align*} (u \cdot v)_\#([\gamma]) = [(u \cdot v)^{-1} \cdot \gamma \cdot (u \cdot v)] = [v^{-1} \cdot u^{-1} \cdot \gamma \cdot u \cdot v]. \end{align*} Now we apply PHL-assoc to regroup this as $[v^{-1} \cdot (u^{-1} \cdot \gamma \cdot u) \cdot v]$. The inner piece $u^{-1} \cdot \gamma \cdot u$ is a loop at $x_1$, and the outer $v^{-1}$ and $v$ represent the change-of-basepoint by $v$ from $x_1$ to $x_2$. That is exactly $v_\#$ applied to the class $[u^{-1} \cdot \gamma \cdot u] = u_\#([\gamma])$: \begin{align*} [v^{-1} \cdot (u^{-1} \cdot \gamma \cdot u) \cdot v] = v_\#([u^{-1} \cdot \gamma \cdot u]) = v_\#(u_\#([\gamma])) = (v_\# \circ u_\#)([\gamma]). \end{align*} **Why reversal?** Think of $u_\#$ as "conjugate by $u$": $\gamma \mapsto u^{-1} \gamma u$. Composing $v_\#$ after $u_\#$ gives $(u^{-1} \gamma u) \mapsto v^{-1}(u^{-1} \gamma u) v = (uv)^{-1} \gamma (uv)$, which is $(u \cdot v)_\#$. In group theory this is the statement that conjugation is an **anti**-representation on the left but a representation on the right; here we recover the functorial statement that $(u \cdot v)_\# = v_\# \circ u_\#$.[/guided]

[step:Prove property (4): naturality with respect to based maps — $(f \circ u)_\# \circ f_* = f_* \circ u_\#$]Let $f: X \to Y$ with $f(x_0) = y_0$ and $f(x_1) = y_1$. Both sides are functions $\pi_1(X, x_0) \to \pi_1(Y, y_1)$; we evaluate them on an arbitrary class $[\gamma] \in \pi_1(X, x_0)$. By [Theorem 1879](/theorems/1879), $f$ commutes with concatenation pointwise on paths: $f \circ (\alpha \cdot \beta) = (f \circ \alpha) \cdot (f \circ \beta)$ as maps $I \to Y$ whenever $\alpha(1) = \beta(0)$, and $f \circ \alpha^{-1} = (f \circ \alpha)^{-1}$. Hence \begin{align*} f \circ (u^{-1} \cdot \gamma \cdot u) = (f \circ u^{-1}) \cdot (f \circ \gamma) \cdot (f \circ u) = (f \circ u)^{-1} \cdot (f \circ \gamma) \cdot (f \circ u). \end{align*} The path $f \circ u$ is a path from $f(x_0) = y_0$ to $f(x_1) = y_1$ in $Y$, so $(f \circ u)_\#$ is a well-defined isomorphism $\pi_1(Y, y_0) \to \pi_1(Y, y_1)$. Now \begin{align*} (f_* \circ u_\#)([\gamma]) &= f_*([u^{-1} \cdot \gamma \cdot u]) = [f \circ (u^{-1} \cdot \gamma \cdot u)] = [(f \circ u)^{-1} \cdot (f \circ \gamma) \cdot (f \circ u)] \\ &= (f \circ u)_\#([f \circ \gamma]) = (f \circ u)_\#(f_*([\gamma])) = ((f \circ u)_\# \circ f_*)([\gamma]). \end{align*} This holds for every class, so the diagram commutes.[/step]

[guided]The assertion is a **naturality square**. In categorical language, if we view $\pi_1$ as a functor from based spaces to groups and the change-of-basepoint maps as isomorphisms, naturality says that push-forward by $f$ commutes with change-of-basepoint. The proof reduces to a single pointwise observation: $f \circ (u^{-1} \cdot \gamma \cdot u) = (f \circ u)^{-1} \cdot (f \circ \gamma) \cdot (f \circ u)$ as functions $I \to Y$. This follows from two facts about the pointwise action of $f$: - **$f$ commutes with concatenation**: $f \circ (\alpha \cdot \beta) = (f \circ \alpha) \cdot (f \circ \beta)$, by the piecewise concatenation formula. - **$f$ commutes with reversal**: $(f \circ \alpha^{-1})(s) = f(\alpha^{-1}(s)) = f(\alpha(1-s)) = (f \circ \alpha)(1 - s) = (f \circ \alpha)^{-1}(s)$. With these in hand, the rest is just chasing the equality through the definitions of $f_*$, $u_\#$, and $(f \circ u)_\#$. Starting on the left: \begin{align*} (f_* \circ u_\#)([\gamma]) = f_*([u^{-1} \cdot \gamma \cdot u]) = [f \circ (u^{-1} \cdot \gamma \cdot u)]. \end{align*} Apply the pointwise identity: \begin{align*} = [(f \circ u)^{-1} \cdot (f \circ \gamma) \cdot (f \circ u)] = (f \circ u)_\#([f \circ \gamma]). \end{align*} The inner class $[f \circ \gamma]$ is $f_*([\gamma])$, so \begin{align*} = (f \circ u)_\#(f_*([\gamma])) = ((f \circ u)_\# \circ f_*)([\gamma]). \end{align*} **Why does this matter?** This is the diagram that allows one to compare fundamental groups **at different basepoints** through maps that change both the space and the basepoint. It is the technical heart of the statement that $\pi_1$, as a functor, is insensitive to the basepoint choice up to canonical isomorphism in a sufficiently coherent way.[/guided]

[step:Prove property (5): when $x_1 = x_0$, $u_\#$ is conjugation by $[u]$] Suppose $x_1 = x_0$. Then $u: x_0 \rightsquigarrow x_0$ is a loop at $x_0$, so $[u] \in \pi_1(X, x_0)$, and $u_\#: \pi_1(X, x_0) \to \pi_1(X, x_0)$ is an automorphism. By definition of the group law on $\pi_1(X, x_0)$, \begin{align*} [u]^{-1} \cdot [\gamma] \cdot [u] = [u^{-1}] \cdot [\gamma] \cdot [u] = [u^{-1} \cdot \gamma] \cdot [u] = [u^{-1} \cdot \gamma \cdot u] = u_\#([\gamma]), \end{align*} where the first equality uses that $[u]^{-1} = [u^{-1}]$ (the inverse in $\pi_1$ is represented by the reversed path — this is the content of the inverse law of [Theorem 1878](/theorems/1878)), and the subsequent equalities use well-definedness of concatenation on classes. So $u_\#([\gamma]) = [u]^{-1} \cdot [\gamma] \cdot [u]$, which is conjugation of $[\gamma]$ by $[u]$ in $\pi_1(X, x_0)$. That is, $u_\# = C_{[u]^{-1}}$ in the usual notation for inner automorphisms (alternatively, $u_\#$ is conjugation by $[u]$ in the convention where conjugation acts by $g \mapsto u^{-1} g u$). In particular, $u_\#$ is an **inner** automorphism. [/step]

[step:Conclude] Steps 1 and 2 produce a well-defined homomorphism $u_\#: \pi_1(X, x_0) \to \pi_1(X, x_1)$; Step 3 shows it is an isomorphism with inverse $(u^{-1})_\#$. Properties (1)–(5) are verified respectively in Steps 4, 5, 6, 7, and 8. This completes the proof. [/step]