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True. Choose the irrational number $\alpha =\sqrt{2}$, and partition the rational numbers intoNo endpoint is rational, so these disjoint sets cover $\mathbb{Q}$. Each is nonempty by the density of the rational numbers, and each is clopen in the subspace topology on $\mathbb{Q}$.

Let two rational numbers be equivalent exactly when they belong to the same $C_n$. The bijectionis well-defined. For every subset $B\subseteq \mathbb{Z}$, the preimage of $h^{-1}(B)$ under the canonical projection of a quotient set is ${\bigcup }_{n\in B}{C}_n$, which is open in $\mathbb{Q}$. Thus every subset of $\mathbb{Q}/\sim$ is open: its quotient topology is the discrete topology. The usual subspace topology on the integers is also the discrete topology, because $\{n\}=\mathbb{Z}\cap (n-1/2,n+1/2)$. Consequently $h$ is a homeomorphism.

Use compatible edge identifications: choose a bijection between the three boundary vertices of $\tau$ and those of $\sigma$, and extend it linearly on each edge. This defines a simplicial homeomorphism of the two boundary circles. This compatibility is the usual meaning of gluing the two boundary triangles; independently chosen edge reversals need not agree at their endpoints. The classification needed here is 3(a)(iv); the printed reference to 3(a)(iii) is a source error.

Delete the two triangular simplexes but retain all their proper faces of a simplex, and identify the two resulting boundary simplicial subcomplexes by this map. The result is a finite abstract simplicial complex $L$: vertices from different sides are identified only in the boundary, and the only shared edges are boundary edges. There is no triangle consisting of all three boundary vertices after the two faces have been removed. Consequently no other simplexes become duplicates. Its geometric realization has the specified quotient topology and is homeomorphic to $Z$.

We check the closed combinatorial surface conditions. Away from the seam, vertex links of a simplex remain cycle graphs. At a boundary vertex of $X$ or $Y$, removing the face changes its link from a cycle graph to a graph path, with endpoints corresponding to its two boundary edges. Gluing the boundaries identifies the two paths' respective endpoints, making one cycle graph. Each seam edge now meets one triangle from $X$ and one from $Y$, and all other edges retain their original two incident triangles. Thus every vertex link is a cycle graph and every simplex lies in a triangle. The simplicial complex is connected, since the two connected punctured surfaces meet along their boundary circle. Hence $L$ is a closed combinatorial surface.

This construction is the connected sum of surfaces $T^2\#{\mathbb{RP}}^2$. For completeness, compute its Euler characteristic directly. Removing a triangular face while retaining its edges and vertices subtracts one, soGluing identifies three vertex pairs and three edge pairs, givingFor the explicit triangulations above, this is also $62-189+126=-1$.

Finally $Z$ is a nonorientable surface. If $Z$ were orientable, its restriction would make $Y$ orientable. The consistent choices of orientation of a simplex on its triangles extend across a capping disc: choose the cap's orientation of a simplex so that the induced directions on the common boundary oppose. Capping $Y$ by $\sigma$ would therefore make the real projective plane orientable, a contradiction. The obstruction to orientability in the real projective plane can also be seen from a Möbius band neighbourhood of a projective line.

By the classification theorem for closed combinatorial surfaces, a nonorientable surface with Euler characteristic $-1$ has $2-k=-1$, hence $k=3$. Therefore

A bijection is a function that is both injective and surjective. Each element of the codomain then has exactly one preimage:This unique-preimage property allows the inverse function $f^{-1}:B\to A$ to be defined on all of $B$.

The inverse factors are generally in the wrong order. For a counterexample, take the bijections $f(x)=x+1$ and $g(x)=2x$, with inverse functions $f^{-1}(x)=x-1$ and $g^{-1}(x)=x/2$. ThenThusFor particular bijections that commute the equality can hold; the claim fails as a universal assertion.

The function composition of two bijections is a bijection. Its inverse function is two-sided, soThis is exactly the defining cancellation property of an inverse function, with the whole function composition regarded as one function.

For the bijections $f(x)=x+1$ and $g(x)=2x$ used above, $f\circ g$ sends $x$ to $2x+1$. Its inverse function sends $x$ to $(x-1)/2$, whereas $f^{-1}\circ g^{-1}$ sends $x$ to $x/2-1$. Hence

Set $y=f(x)$ and cube the equation. Solving for $x^3$ givesConsequently the inverse function isThe value $y=1$ is impossible in the original equation. Conversely, every real $y\ne 1$ gives exactly one real $x$ by the displayed cube root, and this $x$ cannot equal $-1$. Direct substitution verifies both inverse identities. Thus the corrected bijection is $f:\mathbb{R}\setminus \{-1\}\to \mathbb{R}\setminus \{1\}$.

List the $n$ elements of the finite set $A$. A subset is determined by $n$ membership decisions, each with two possibilities. This is a bijection from the power set to the set $\{0,1{\}}^n$, soThe empty set case is included: its power set contains the single empty subset.

Swap the components of every ordered pair using a list comprehension:

invert :: [(a, a)] -> [(a, a)]
invert r = [(y, x) | (x, y) <- r]

Thus $\boxed{\mathrm{invert}(R)=R^{-1}}$, the converse relation. Swapping is a bijection, since doing it twice restores the original pair. Therefore distinct input pairs remain distinct; no equality constraint or duplicate removal is necessary.

If $f(x)=f(y)$, applying the function $f$ gives $f(f(x))=f(f(y))$. Since $f\circ f={\mathrm{id}}_A$, this implies $x=y$, proving injectivity. Also every $y\in A$ is the image of $f(y)$, because $f(f(y))=y$, proving surjectivity.

Thus $f$ is a bijection and its inverse function is itself:Such a function is an involution.

With the convention $\mathbb{N}=\{0,1,2,\dots \}$, the even inputs $2j$ map to $j$ for $j\ge 0$, while the odd inputs $2j+1$ map to $-(j+1)$ for $j\ge 0$. These two images of a function are disjoint and together equal $\mathbb{Z}$. Within either class there are no repeated outputs. More explicitly the inverse function isConsequently $f$ is both injective and surjective, hence a bijection. If instead $\mathbb{N}=\{1,2,\dots \}$, the input zero is absent: $f$ remains injective but its image of a function is $\mathbb{Z}\setminus \{0\}$, so it is not onto. The printed paper does not specify this convention; the zero-based convention gives the apparent intended bijection.

Put one object into each box first. The remaining $n-k$ objects can be distributed without restrictions, so stars and bars givesEquivalently, the substitution $u_i=x_i-1$ gives a bijection between positive integer solutions of ${\sum }_i{x}_i=n$ and nonnegative integer solutions of ${\sum }_i{u}_i=n-k$.

A bijection between finite sets requires equal cardinality. When $m=n$, list the elements of $A$ and choose their distinct images in $n,n-1,\dots ,1$ ways. ThusThe factorial convention $0!=1$ includes the unique bijection from the empty set to itself.

A function $f:A\to B$ is onto, or a surjective function, when every $b\in B$ has a preimage: $\forall b\in B\exists a\in A:f(a)=b$. It is one-to-one, or an injective function, when $f(a)=f(a^{\prime })$ implies $a=a^{\prime }$ for all $a,a^{\prime }\in A$. It is bijective, a bijection, when both properties hold.

An inverse function is a function $f^{-1}:B\to A$ satisfyingFor a bijection, $f^{-1}(b)$ is the unique $a$ with $f(a)=b$: existence follows from surjectivity and uniqueness from injectivity. This is a two-sided inverse function, rather than just a left or right inverse.

Let $h:B\to A$ be a two-sided inverse function of $f$. If $f(a)=f(a^{\prime })$, apply $h$ and use $h\circ f={\mathrm{id}}_A$ to obtain $a=a^{\prime }$, proving injectivity. For any $b\in B$, the other identity gives $b=f(h(b))$, proving surjectivity. Therefore $f$ is a bijection. Both inverse identities matter: a one-sided inverse alone would establish only the corresponding one of these two properties.

Separate integer partitions according to whether they contain a part equal to $i$. Removing one such part is a bijection to partitions of $n-i$ whose parts are at most $i$. The alternatives are disjoint, givingUse $p(i,0)=1$, because the empty integer partition is counted once, and $p(0,n)=0$ for $n>0$, with $p(i,n)=0$ for $n<0$. This Oberon implementation avoids negative recursive arguments:The precondition is $i,n\ge 0$. Each recursive call reduces $i+n$, proving algorithm termination. Every positive part of a partition of $n$ is at most $n$, so the required one-argument count is $P(n,n)$. Counts require a sufficiently wide integer representation.

One example is the identity function, $f(n)=n$. A second is the permutationBoth are bijections of $\mathbb{N}$, whichever convention includes or excludes zero. For the second, applying $f$ three times returns every input to itself. Thus $f^3=\mathrm{id}$, and composing with the inverse function gives $f^2=f^{-1}$. This permutation cube identity criterion also shows why three-cycles work whereas a nontrivial two-cycle would not.

A derangement of a finite set $A$ is a permutation $\sigma :A\to A$ with no fixed point: $\sigma (a)\ne a$ for every $a\in A$. Thus every element is moved, although the permutation remains a bijection.

A subset is determined uniquely by deciding, for each of the $n$ elements, whether to include it. Each decision has two possibilities. By the rule of product, there are $2\cdot 2\cdots 2=2^n$ decision strings. The correspondence between such binary strings and subsets is a bijection, so $|\mathcal{P}(\mathcal{U})|=2^n$. The all-zero string corresponds to the empty subset.

An allocation is a tuple $(x_1,\dots ,x_k)$ of nonnegative integers with ${\sum }_j{x}_j=n$, where $x_j$ is the number in labelled box $j$. Empty boxes are allowed. Encode the allocation by $n$ stars and $k-1$ bars: put $x_1$ stars before the first bar, $x_2$ between the first two bars, and so on. Adjacent bars represent an empty intermediate box; a bar at an end represents an empty end box.

Reading the group sizes recovers exactly one allocation from every such string, so this is a bijection. There are $n+k-1$ positions and we choose the $k-1$ bar positions. Hence stars and bars givesNo extra factor for ordering the objects is present, because the objects are indistinguishable. For $k=1$ the string has no bars and gives the single allocation of all objects to that box; for $n=0$ all boxes are empty, also a single allocation.