Definition 1.4.6.1. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. We will say that $f$ is an *isomorphism* if the homotopy class $[f]$ is an isomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. We will say that two objects $X,Y \in \operatorname{\mathcal{C}}$ are *isomorphic* if there exists an isomorphism from $X$ to $Y$ (that is, if $X$ and $Y$ are isomorphic as objects of the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$).

### 1.4.6 Isomorphisms

Recall that a morphism $f: X \rightarrow Y$ in a category $\operatorname{\mathcal{C}}$ is an *isomorphism* if there exists a morphism $g: Y \rightarrow X$ satisfying $f \circ g = \operatorname{id}_{Y}$ and $g \circ f = \operatorname{id}_ X$. This notion has an $\infty $-categorical analogue:

Example 1.4.6.2. Let $\operatorname{\mathcal{C}}$ be an ordinary category. Then a morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ is an isomorphism if and only if it is an isomorphism when regarded as a morphism of the $\infty $-category $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$.

Remark 1.4.6.3 (Two-out-of-three). Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be morphisms in an $\infty $-category $\operatorname{\mathcal{C}}$ and let $h$ be a composition of $f$ and $g$. If any two of the morphisms $f$, $g$, and $h$ is an isomorphism, then so is the third.

Definition 1.4.6.4. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and suppose we are given a pair of morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow X$ in $\operatorname{\mathcal{C}}$. We say that $g$ is a *left homotopy inverse* of $f$ if the identity morphism $\operatorname{id}_{X}$ is a composition of $f$ and $g$: that is, if we have an equality $[\operatorname{id}_ X] = [g] \circ [f]$ in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. We say that $g$ is a *right homotopy inverse* of $f$ if the identity morphism $\operatorname{id}_{Y}$ is a composition of $g$ and $f$: that is, if we have an equality $[ \operatorname{id}_ Y ] = [f] \circ [g]$ in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. We will say that $g$ is a *homotopy inverse* of $f$ if it is both a left and a right homotopy inverse of $f$.

Remark 1.4.6.5. Let $f: X \rightarrow Y$ and $g: Y \rightarrow X$ be morphisms in an $\infty $-category $\operatorname{\mathcal{C}}$. Then the condition that $g$ is a left homotopy inverse (right homotopy inverse, homotopy inverse) to $f$ depends only on the homotopy classes $[f]$ and $[g]$.

Remark 1.4.6.6. Let $f: X \rightarrow Y$ and $g: Y \rightarrow X$ be morphisms in an $\infty $-category $\operatorname{\mathcal{C}}$. Then $g$ is left homotopy inverse to $f$ if and only if $f$ is right homotopy inverse to $g$. Both of these conditions are equivalent to the existence of a $2$-simplex $\sigma $ of $\operatorname{\mathcal{C}}$ satisfying $d^{2}_0(\sigma ) = g$, $d^{2}_1(\sigma ) = \operatorname{id}_{X}$, and $d^{2}_2(\sigma ) = f$, as depicted in the diagram

Remark 1.4.6.7. Let $f: X \rightarrow Y$ be a morphism in an $\infty $-category $\operatorname{\mathcal{C}}$. Suppose that $f$ admits a left homotopy inverse $g$ and a right homotopy inverse $h$. Then $g$ and $h$ are homotopic: this follows from the calculation

It follows that both $g$ and $h$ are homotopy inverse to $f$.

Remark 1.4.6.8. Let $f: X \rightarrow Y$ be a morphism in the $\infty $-category $\operatorname{\mathcal{C}}$. It follows from Remark 1.4.6.7 that the following conditions are equivalent:

- $(1)$
The morphism $f$ is an isomorphism.

- $(2)$
The morphism $f$ admits a homotopy inverse $g$.

- $(3)$
The morphism $f$ admits both left and right homotopy inverses.

In this case, the morphism $g$ is uniquely determined up to homotopy; moreover, any left or right homotopy inverse of $f$ is homotopic to $g$. We will sometimes abuse notation by writing $f^{-1}$ to denote a homotopy inverse to $f$.

Warning 1.4.6.9. Let $f: X \rightarrow Y$ be a morphism in an $\infty $-category $\operatorname{\mathcal{C}}$, and suppose that $g,h: Y \rightarrow X$ are left homotopy inverses to $f$. If $f$ does not admit a right homotopy inverse, then $g$ and $h$ need not be homotopic.

Proposition 1.4.6.10. Let $\operatorname{\mathcal{C}}$ be a Kan complex. Then every morphism in $\operatorname{\mathcal{C}}$ is an isomorphism.

Remark 1.4.6.11. We will see later that the converse to Proposition 1.4.6.10 is also true: if $\operatorname{\mathcal{C}}$ is an $\infty $-category in which every morphism is an isomorphism, then $\operatorname{\mathcal{C}}$ is a Kan complex (Proposition 4.4.2.1).

**Proof of Proposition 1.4.6.10.**
Let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$. Then the tuple $(\bullet , \operatorname{id}_{X}, f)$ determines a map of simplicial sets $\sigma _0: \Lambda ^{2}_{0} \rightarrow \operatorname{\mathcal{C}}$ (Proposition 1.2.4.7), which we depict as

If $\operatorname{\mathcal{C}}$ is a Kan complex, then we can extend $\sigma _0$ to a $2$-simplex $\sigma $ of $\operatorname{\mathcal{C}}$. Then $\sigma $ exhibits the morphism $g = d^{2}_0(\sigma )$ as a left homotopy inverse to $f$. A similar argument shows that $f$ admits a right homotopy inverse, so that $f$ is an isomorphism by virtue of Remark 1.4.6.8. $\square$

Definition 1.4.6.12 (The Fundamental Groupoid of a Kan Complex). Let $X$ be a Kan complex. It follows from Proposition 1.4.6.10 that the homotopy category $\mathrm{h} \mathit{X}$ of Definition 1.4.5.3 is a groupoid. We will denote this groupoid by $\pi _{\leq 1}(X)$ and refer to it as the *fundamental groupoid* of $X$.

Remark 1.4.6.13. Let $X$ be a Kan complex. By construction, the objects of the fundamental groupoid $\pi _{\leq 1}(X)$ are the vertices of $X$, and a pair of vertices $x,y \in X$ are isomorphic in $\pi _{\leq 1}( X )$ if and only if there exists an edge $e: x \rightarrow y$ in $X$. Applying Proposition 1.2.5.10, we deduce that $x,y \in X$ are isomorphic if and only if they belong to the same connected component of $X$. In other words, we have a canonical bijection

Example 1.4.6.14. Let $X$ be a topological space. Then the singular simplicial set $\operatorname{Sing}_{\bullet }(X)$ is a Kan complex (Proposition 1.2.5.8), and its fundamental groupoid $\pi _{\leq 1}( \operatorname{Sing}_{\bullet }(X) )$ can be identified with the usual fundamental groupoid $\pi _{\leq 1}(X)$ of the topological space $X$ (where objects are the points of $X$ and morphisms are given by homotopy classes of paths in $X$).