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The unit circle is the inverse image of the closed set $\{1\}$ under the continuous map $(x,y)\mapsto x^2+y^2$, so it is closed. It is also bounded, since every point has Euclidean norm one. The Heine-Borel theorem gives compactness.

A metric space is path-connected if every pair $x,y$ can be joined by a path: a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.The path must remain in $X$, including at its endpoints.

For $z\in \mathbb{C}\setminus [1,\infty )$, the number $w=1-z$ avoids the nonpositive real axis. Define the branch of the complex logarithm byHere the principal argument is selected as a continuous map on that slit domain, so this is a holomorphic function, with derivative $-1/(1-z)$ and $L(1)=0$. For example, locally writing $w=\rho e^{i\varphi }$ gives $d(\log \rho +i\varphi )=dw/w$, verifying the complex derivative. The normalization also makes this branch of the complex logarithm unique: any two continuous choices differ by a constant integer multiple of $2\pi i$ on the connected slit domain, and their values at $w=1$ agree.

To evaluate the corresponding analytic branch of a square root on the unit circle, factorFor $0<\theta <2\pi$, we have $\sin (\theta /2)>0$ and $-\pi /2<\theta /2-\pi /2<\pi /2$. Thus this factorization already uses the selected principal argument, with no hidden change by $2\pi$. It follows thatHalving this expression and applying the exponential function givesSince $\sqrt{2}{e}^{-i\pi /4}=1-i$, the desired branch value isFor instance, $\theta =\pi$ gives $\sqrt{2}>0$, a quick check that the sign agrees with the normalization.

A metric space $(X,d)$ is a complete metric space if every Cauchy sequence in $X$ converges to a point of $X$. Explicitly, the hypothesis isand the conclusion is that some $x\in X$ satisfies $d(x_n,x)\to 0$.

It is a sequentially compact space if every sequence in $X$ has a convergent subsequence whose limit of a sequence belongs to $X$.

It is a connected space if there are no disjoint nonempty open subsets $U,V$ of $X$ with $X=U\cup V$. Here openness refers to the metric topology of $X$.

It is a path-connected space if, for every $x,y\in X$, there is a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$. Such a map is a path. Every path-connected space is a connected space: a separation would pull back along a path joining its two sides to a separation of the interval $[0,1]$.

Both properties are preserved by a continuous map.

For sequential compactness, take any sequence $(y_n)$ in $f(X)$ and choose $x_n\in X$ with $f(x_n)=y_n$. If $X$ is sequentially compact, some subsequence satisfies $x_{n_k}\to x\in X$. By continuity, $y_{n_k}=f(x_{n_k})\to f(x)\in f(X)$. Hence $f(X)$ is a sequentially compact space.

For connectedness, suppose instead that $f(X)=U\cup V$ is a separation into disjoint nonempty relatively open sets. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint nonempty open sets of $X$ whose union is $X$, contradicting connectedness. Therefore $f(X)$ is a connected space.

Take distinct classes $p(x)\ne p(y)$. The points are not related by $R$, so by the defining negation there is a Hausdorff space $Z$ and a continuous map $f:X\to Z$ with $f(x)\ne f(y)$. Choose disjoint open neighborhoods $U$ and $V$ of those two values. Their inverse images $A=f^{-1}(U)$ and $C=f^{-1}(V)$ are disjoint open subsets of $X$ and are saturated by part (ii).

Hence $p^{-1}(p(A))=A$ and $p^{-1}(p(C))=C$, making $p(A)$ and $p(C)$ open in the quotient topology. They contain the chosen classes and are disjoint: an equivalence class meeting both would contradict saturation and $A\cap C=\varnothing$. Thus $X/R$ is Hausdorff. This quotient is the Hausdorff reflection by continuous maps; every continuous map from $X$ to a Hausdorff space factors uniquely through it.

For each continuous map to a Hausdorff target, equality of its values is reflexive, symmetric and transitive. The relation $R$ requires all these equalities simultaneously. Explicitly, $f(x)=f(x)$ gives reflexivity; $f(x)=f(y)$ implies $f(y)=f(x)$; and $f(x)=f(y)=f(z)$ gives $f(x)=f(z)$. These arguments apply to every map in the defining family, proving $R$ is an equivalence relation.

The claim is false for an arbitrary codomain $Y$. Let $X=[0,1]$ with its usual topology, let $Y$ be the Sierpiński space with open sets $\varnothing ,\{1\},Y$, and define $f(0)=0$, $f(x)=1$ for $x>0$. This is a continuous map: the only nontrivial open-set preimage is $f^{-1}(\{1\})=(0,1]$, which is relatively open in $X$.

The sets $K_n=[0,1/n]$ are nonempty, nested and compact. ButThus the extra generality in the printed codomain changes the assertion materially. This is the failure of nested compact image intersection without closed fibers.

If the codomain is a T1 space, the continuous images of nested compact intersections theorem does give the desired equality. The forward inclusion holds for every map. For the reverse, take $y\in {\bigcap }_{n}f(K_n)$ and set $F_n=K_n\cap f^{-1}(\{y\})$. The fibers are closed by continuity and the T1 property, so the $F_n$ are nested nonempty closed subsets of the compact set $K_1$. The finite intersection property gives $x\in {\bigcap }_n{F}_n$, with $f(x)=y$. It is enough that these particular fibers be closed in $K_1$; a Hausdorff codomain, as in the native self-map formulation, is a familiar sufficient assumption.

A subset $A$ is connected if its subspace topology admits no separation into two disjoint nonempty relatively open subsets. Equivalently, every continuous map $A\to \{0,1\}$, with the target discrete, is constant: a nonconstant map gives the separation by its two fibers, and a separation defines such a map.

A continuous map $f:X\to Y$ is one for which $f^{-1}(U)$ is open in $X$ whenever $U$ is open in $Y$. This defines continuity using the two topologies, without a metric or a coordinate description.

A basis of a topology is a collection $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. Equivalently, for every open $U$ and $y\in U$ there is $B\in \mathcal{B}$ with $y\in B\subseteq U$.

A map $f:X\to Y$ is continuous exactly when $f^{-1}(B)$ is open in $X$ for every basic set $B$ of $Y$. Necessity follows from the definition of a continuous map. For sufficiency, express an arbitrary open set as a union of basic sets and use preservation of unions by inverse images.

True. The Hausdorff assumption makes $\{y\}$ closed in $Y$. Its inverse image under the continuous map $f$ is closed in the compact space $X$, and a closed subset of a compact space is compact. This is the compactness of the fiber with its subspace topology.

The image $f(K)$ is compact: pull any open cover back by the continuous map, use a finite subcover of $K$, then map forward. For a general topological target, closedness is not guaranteed; a continuous image into the indiscrete two-point space can be a nonclosed singleton. No connectedness assertion follows merely from compactness of $K$.

If $Y$ is metric, $f(K)$ is compact, therefore closed in $Y$, bounded and totally bounded. It is also complete in its induced metric and sequentially compact. For the latter properties, compact metric spaces have a convergent subsequence of every sequence; a Cauchy sequence with one convergent subsequence converges to that same limit. Total boundedness follows by selecting finitely many balls of each prescribed radius. These are properties of the image itself, irrespective of whether $Y$ is complete.

A homotopy from $f_0:Y\to X$ to $f_1:Y\to X$ is a continuous map $H:Y\times [0,1]\to X$ with $H(y,0)=f_0(y)$ and $H(y,1)=f_1(y)$ for every $y$. A covering map $p:\tilde{X}\to X$ is a surjective continuous map for which every $x\in X$ has an open neighborhood $U$ whose inverse image is a disjoint union of open sets, each mapped homeomorphically onto $U$. A lift of $f:Y\to X$ is a continuous map $\tilde{f}:Y\to \tilde{X}$ satisfying $p\circ \tilde{f}=f$.

The path lifting theorem says that a path $\gamma :[0,1]\to X$ and a point ${\tilde{x}}_0\in p^{-1}(\gamma (0))$ determine a unique path $\tilde{\gamma }:[0,1]\to \tilde{X}$ with $p\tilde{\gamma }=\gamma$ and $\tilde{\gamma }(0)={\tilde{x}}_0$.

On $S^1\setminus \{1\}$ there is a continuous branch ${\mathrm{Arg}}_{(0,2\pi )}$ of the complex argument, taking values in $(0,2\pi )$. If $f$ avoids $1$, set $\theta (z)={\mathrm{Arg}}_{(0,2\pi )}f(z)$. Then the explicit homotopystarts at $f$ and ends at the constant map $-1$. It is continuous because the chosen argument branch is continuous on the whole image. More generally, a map omitting any $w\in S^1$ is null-homotopic: apply the formula to $\overline{w}f$ and multiply the resulting homotopy by $w$.

For the based injective map, use the standard covering map $E(t)=e^{2\pi it}$. Lift the path $t\mapsto g(E(t))$ on $[0,1]$ to a real-valued path $G$ with $G(0)=0$. Since $g(1)=1$, its endpoint $d=G(1)$ is an integer. On $[0,1)$, $G$ is injective: equality $G(s)=G(t)$ implies $g(E(s))=g(E(t))$, then $E(s)=E(t)$ by injectivity, and hence $s=t$ in this half-open interval. A continuous injective real-valued function on an interval is strictly monotone.

If $G$ is increasing, $d>0$. It cannot be at least $2$, because the intermediate value theorem would produce $t\in (0,1)$ with $G(t)=1$, giving $g(E(t))=g(1)$ while $E(t)\ne 1$. Thus $d=1$. If $G$ is decreasing, the same reasoning gives $d=-1$. Now defineAt $t=0$ and $t=1$ this has the same value $1$ for every $s$, since $d$ is an integer. Therefore the formula descends continuously from $[0,1]\times [0,1]$ to $S^1\times [0,1]$. At $s=0$ it is $g$; at $s=1$ it is $z\mapsto z^d$. Hence $g$ is homotopic to the identity or to complex conjugation, with degree $1$ or $-1$. This is the key property of an injective self-map of a circle.

Finally, let $h$ be any continuous injective circle self-map. Normalize it to $g(z)=\overline{h(1)}h(z)$, which is injective and fixes $1$. If $h$ were not surjective, it would omit some circle point, so the explicit missing-point construction would make $h$ null-homotopic. Multiplying this homotopy by $\overline{h(1)}$ would make $g$ null-homotopic too. But $g$ is homotopic to the identity or conjugation, neither of which is null-homotopic by the permitted fact. This contradiction proves every continuous injective circle self-map is surjective. The lifting argument did not assume surjectivity, so the reasoning is not circular.

Let $i:A\hookrightarrow X$ be the inclusion map and $r:X\to A$ the retraction. Since $X$ is contractible, choose a contraction of a topological spaceThe function compositionis a continuous map. Its endpoints satisfyConsequently $K$ is a homotopy from the identity map of $A$ to a constant function. This provesThe contracting point of $X$ need not belong to $A$: applying the retraction sends it to the contracting point $r(x_0)$ of $A$. This is the retract of a contractible space principle.

Yes. Take $X=S^1$, the circle, and $A=\{1\}\subset S^1$. The constant function $r(z)=1$ is a continuous map and satisfies $ri={\mathrm{id}}_A$, so $A$ is a retract of $X$.

However, their fundamental groups areA homotopy equivalence between path-connected spaces induces a group isomorphism of their fundamental groups, allowing the usual change of base point. The infinite cyclic group is not the trivial group. Therefore a retract need not have the homotopy type of its ambient space.

Two continuous maps $f_0,f_1:X\to Y$ are homotopic if there is a continuous mapHere the Cartesian product has the product topology. Such an $H$ is a homotopy, written $f_0\simeq f_1$.

The topological spaces $X,Y$ are homotopy equivalent if there are continuous maps $f:X\to Y$ and $g:Y\to X$ such thatEach is then a homotopy inverse of the other.

For $A\subseteq X$ with its subspace topology with inclusion map $i:A\hookrightarrow X$, a homotopy retract, in the course's convention, is a continuous map $r:X\to A$ withThus $r$ is an actual retraction, and $i,r$ give a homotopy equivalence. Requiring the latter homotopy to fix $A$ pointwise at all times gives the stronger condition of a deformation retraction as used in this wiki. The course's definition of homotopy retract requires no such restriction on that homotopy.

Identify the bottom $j(X)=q(X\times \{0\})$ with $X$; its natural map $j$ is a topological embedding. To verify this directly, an open subset $O\subseteq X$ gives the saturated open subset $O\times [0,1/2)$ of the space before taking the quotient topology; its image is open in $M(f)$ and meets the bottom in $j(O)$.

Suppose first that there is a retraction $R:M(f)\to j(X)$. Define the continuous mapThenis a homotopy with $L(x,0)=x$ and $L(x,1)=g(f(x))$. Hence $gf\simeq {\mathrm{id}}_X$.

Conversely, suppose a continuous map $g:Y\to X$ has $gf\simeq {\mathrm{id}}_X$. Reverse the parameter if necessary to choose a homotopy $L$ with $L(x,0)=x$ and $L(x,1)=gf(x)$. SetAt the glued end, $j(L(x,1))=j(gf(x))$, exactly the value prescribed at $i(f(x))$. The universal property of the quotient topology therefore gives a continuous map $R:M(f)\to j(X)$. At the bottom, $R(j(x))=j(x)$, proving that $R$ is a retraction. We have established the mapping-cylinder bottom retraction criterion:

DefineThe two formulas agree on the attached end, so the universal property of the quotient topology makes $r$ a continuous map. Also $ri={\mathrm{id}}_Y$.

Slide each cylinder point towards its attached end usingAt $t=1$ these formulas both give $i(f(x))$, so they define a homotopy on the mapping cylinder. For continuity, the product of a quotient map with the unit interval is a quotient map; the formulas before taking the quotient topology are continuous maps and agree on every identified pair. ThusMoreover $Y$ stays fixed throughout. We have obtained the stronger conclusionIn particular $r$ is the requested homotopy retract.

Writefor the quotient map defining the mapping cylinder, and put $j(x)=q(x,0)$. Since the equivalence relation only attaches points at height $1$ to $Y$, distinct points at height $0$ remain distinct. Thus $j$ is an injective continuous map.

Define the homotopyIt is a continuous map, with $H(x,0)=q(x,1)=i(f(x))$ and $H(x,1)=j(x)$. HenceEven if $f$ identifies many points, the free end of the mapping cylinder retains a separate copy of every point of $X$.