Jekyll2023-05-08T00:50:41+00:00http://www.topos.place/feed.xmlWelcome to Topos.PlacePersonal webpage and math blog of Charlie ConneenCharlie ConneenAn interesting calculus problem2023-05-06T00:00:00+00:002023-05-06T00:00:00+00:00http://www.topos.place/integration-by-substitution-using-complex-geometry<style>
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<h3 id="and-a-proof-using-complex-geometry">…and a proof using complex geometry</h3>
<p>Note for the reader: the first two sections of this post are accessible to anyone with only a little background in integral calculus and some familiarity with the complex numbers. However, the last section requires at least some familiarity with complex manifolds. This blog post was originally going to only include the proof in the last section, but I decided to look into the calculus content of the result and found it interesting and highly accessible, so I’ve included it here.</p>
<h2 id="the-result-and-proof-1-using-calculus">The result, and proof #1, using calculus</h2>
<p>Every calculus student has experienced the dread of computing a tricky integral involving trigonometric functions. For example, consider the following integral, which a particularly cruel professor might put on an assignment:</p>
<div style="text-align: center; padding-bottom: 10px">$\displaystyle\int \frac1{\cos(x)+2}\mathrm{d}x$</div>
<p>This is surprisingly nontrivial! One solution is to use the following trigonometric facts:</p>
<div style="text-align: center; padding-bottom: 15px">
$\displaystyle\cos(x) = 2\cos^2\left(\frac x2\right) - 1$
</div>
<div style="text-align: center; padding-bottom: 15px">
$\displaystyle\cos(x) = 2\cos^2\left(\frac x2\right) - 1$
</div>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle\sec^2\left(\frac x2\right) = \tan^2\left(\frac x2\right) + 1$
</div>
<p>and rewrite the integral as</p>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle\int \frac1{\cos(x)+2}\mathrm{d}x = \int \frac{\sec^2\left(\frac x2\right)}{\tan^2\left(\frac x2\right)+3}\mathrm{d}x,$
</div>
<p>leaving it to the reader to verify this equality, using the trig facts as outlined. From here, an integral substitution using $u=\tan\left(\frac x2\right)$ allows us to rewrite this as</p>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle\int \frac{2}{u^2+3}\mathrm{d}u.$
</div>
<p><strong>Exercise for calculus enjoyers:</strong> finish the above integral.</p>
<div style="border: solid; padding: 4px">
<details>
<summary>Big hint</summary>
Write
<div style="text-align: center; padding-bottom: 15px">
$\displaystyle\frac{2}{u^2+3} = \frac 23\cdot\frac 1{(\frac{u}{\sqrt{3}})^2+1}$
</div>
and use the fact that $\int \frac{1}{1+x^2}\mathrm{d}x=\arctan(x)+C$.
</details>
</div>
<p>Computations aside, the main takeaway of the above example is this: we were able (in this case, at least) to compute the integral of a rational trig function via substitution, rewriting it as a rational function with no reference to trigonometric functions.</p>
<p>Here are two other examples of (properly nasty) rational trig function integrals, and their corresponding (also nasty) rational functions after an appropriate substitution:</p>
<div style="text-align: center; padding-bottom: 15px">
$\displaystyle\int\frac{1}{\sin(x)+\cos(x)}\mathrm{d}x = \int\frac{-2}{u^2-2u-1}\mathrm{d}u$
</div>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle\int\frac{\sin^2(x)}{\sin(x)+\cos(x)}\mathrm{d}x = \int\frac{-8u^2}{(u^2+1)^2(u^2-2u+1)}\mathrm{d}u$
</div>
<p>Hopefully this has convinced the reader that it’s reasonably likely this process can be done for any rational trig function, although it may be completely unclear how this is achieved. (Of course, these rewritten forms of the trig integrals are equally unapproachable calculus problems as before they were rewritten, but our ability to do this rewriting is an interesting phenomenon nonetheless.)</p>
<p>What I haven’t told you is that all of the above integrals use the exact same substitution, namely, $u=\tan(\frac x2)$.</p>
<p><strong>Exercise for the skeptical:</strong> Verify that if $u=\tan(\frac x2)$, then $\sin(x)=\frac{2u}{1+u^2}$, $\cos(x)=\frac{1-u^2}{1+u^2}$, and $\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{2}{1+u^2}$.</p>
<p>With this exercise out of the way, we are able to see the main result: if $f(x,y)$ is a rational function in two variables, then $\int f(\sin(x),\cos(x))\mathrm{d}x$ can be rewritten as an integral of the form $\int g(u)\mathrm{d}u$, where $g(u)$ is a rational function. Namely, the above exercise tells us:</p>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle\int f(\sin(x),\cos(x))\mathrm{d}x = \int f\left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\frac{1+u^2}2\mathrm{d}u,$
</div>
<p>which is the integral of a rational function after some minimal rearrangement.</p>
<p>However, this is an unsatisfying solution to this problem: the function $\tan(\frac x2)$ was presented from the aether, and just so happened to satisfy the substitution requirements. Plus, the fact that this substitution even works requires too much manual computation using trig facts to be enjoyable. We will now go through two other proofs that there is such an integral substitution, both of which will seem more natural. In the first of these, we will present a proof using only calculus and a little bit of knowledge of the complex numbers. In the second, we will take we will take the “higher” road, and look to complex geometry to answer our questions.</p>
<h2 id="proof-2-also-calculus-but-much-cleaner">Proof #2, also calculus, but much cleaner</h2>
<p>Again let $f(\sin(x),\cos(x))$ be a rational trig function. To reinterpret the functions $\sin(x)$ and $\cos(x)$ in a way more conducive to substitution, we use the classical identity</p>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle e^{ix} = \cos(x) + i\sin(x)$
</div>
<p>to rewrite $\sin$ and $\cos$ as</p>
<div style="text-align: center; padding-bottom: 15px">
$\displaystyle \sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$
</div>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle \cos(x) = \frac{e^{ix}+e^{-ix}}{2}$
</div>
<p>Using this, we can rewrite</p>
<div style="text-align: center">
$\displaystyle f(\sin(x),\cos(x)) = f\left(\frac{e^{ix}-e^{-ix}}{2i},\frac{e^{ix}+e^{-ix}}{2}\right)$
</div>
<p>Now an appropriate substitution is clear: let $u=e^{ix}$. Then $\frac{\mathrm du}{\mathrm dx} = ie^{ix}= i\cdot u$, hence</p>
<div style="text-align: center">
$\displaystyle \int f(\sin(x),\cos(x))\mathrm dx = \int f\left(\frac{u-u^{-1}}{2i},\frac{u+u^{-1}}{2}\right)\cdot -iu^{-1}\mathrm du$
</div>
<p>which rewrites the integral as a rational function of $u$, after an appropriate clearing of negative exponents. This is likely the “easiest” of the methods which will be presented here, but it does yield an integral with complex coefficients in general, which may be undesirable if you’re trying to use this method on a calculus homework.</p>
<h2 id="proof-3-via-complex-geometry">Proof #3, via complex geometry</h2>
<p>This proof is perhaps the least satisfying for those who wish to compute things, as it merely proves that there exists a substitution; it doesn’t give an explicit method to do so (at least, on the surface), as the previous proofs do. However, there are two bits of good news:</p>
<ol>
<li>If you are computationally minded, this method (in theory) provides a way to obtain more methods of substitution.</li>
<li>for those who do not enjoy computing things, this will (I hope!) be the most satisfying of the three approaches, because of how little work needs to be done.</li>
</ol>
<p>Let $\mathscr{O}(\mathbb{C})$ denote the ring of entire functions, and let $A=\mathbb{C}[\sin(z),\cos(z)]$ denote the smallest subring containing the constant functions (regarded as elements of $\mathbb{C}$), along with the functions $\sin(z)$ and $\cos(z)$. Now observe that $A\cong\mathbb{C}[x_0,x_1]/(x_0^2+x_1^2-1)$ (interpreting the image of $x_0,x_1$ under this quotient as $\sin$ and $\cos$, respectively). Then we can projectivize the polynomial generating that ideal:</p>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle f(x_0,x_1)=x_0^2+x_1^2-1\;\rightsquigarrow\; F(x_0,x_1,x_2) = x_0^2+x_1^2-x_2^2$
</div>
<p>This allows us to construct the corresponding projective variety:</p>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle Z(F) = \{[X_0:X_1:X_2]\in\mathbb{P}^2\mid F(X_0,X_1,X_2)=0\}\subseteq\mathbb{P}^2,$
</div>
<p>where $\mathbb{P}^2$ denotes the complex projective plane. By construction, we may identify meromorphic $1$-forms on this space with those of the form $g(z)\mathrm{d}z$ where $g$ is a rational trig function. It’s also a straightforward exercise to show that this curve is genus zero. Since the only projective curve of genus zero over the complex numbers up to (analytic) isomorphism is $\mathbb{P}^1$, we can equip ourselves with an analytic isomorphism $\varphi\colon \mathbb{P}^1\cong Z(F)$.</p>
<div style="border: solid; padding:4px">
<details>
<summary>Aside, for those curious about genera</summary>
The fact that this curve is genus zero is really a consequence of the "Plücker formula" which states that, for a smooth projective curve of degree $d$ (meaning the defining equation is a degree $d$ homogeneous polynomial), the genus of the curve is $g=\frac12(d-1)(d-2)$. The degree of the curve in question is $2$, so its genus is indeed zero. A good reference for this fact is Rick Miranda's book "Algebraic Curves and Riemann Surfaces" pp. 143-145. If the reader is instead curious about why the only genus zero curve is the Riemann sphere, the same reference can answer this, although this is a less technical result.
</details>
</div>
<p>Now let $\omega = g(z)\mathrm{d}z$ be a meromorphic $1$-form on $Z(F)$. Then $\omega$ is pulled back via the isomorphism $\varphi$ to a meromorphic $1$-form $\varphi^\ast(\omega)$ on $\mathbb{P}^1$. Since the only meromorphic $1$-forms on the Riemann sphere are of the form $\frac{p(z)}{q(z)}\mathrm{d}z$ where $p,q$ are polynomials, this exhibits the integral substitution!</p>
<h3 id="some-closing-remarksneat-facts">Some closing remarks/neat facts</h3>
<ol>
<li>Again consider the ring $A$ as defined above, and let $K=\mathrm{Frac}(A)$ be its fraction field. Similarly, let $L=\mathrm{Frac}(\mathbb{C}[x_0,x_1]/(x_0^2+x_1^2-1))$. The integral substitution in the first proof shows that the element $\tan\left(\frac x2\right)=\frac{\sin(x)}{1+\cos(x)}\in K$ generates the field extension $K/\mathbb{C}$. Combining this with the isomorphism $A\cong\mathbb{C}[x_0,x_1]/(x_0^2+x_1^2-1)$ gives that $\frac{x_0}{1+x_1}\in L$ generates the field extension $L/\mathbb{C}$, which is a somewhat funny way of arriving at this fact.</li>
<li>The same logic applies to the substitution in proof #2 as observed in proof #1: the substitution $u=e^{ix}$ being valid means that $\cos(x)+i\sin(x)\in K$ generates the extension $K/\mathbb{C}$, and therefore $x_1+ix_0\in L$ generates the extension $L/\mathbb{C}$ (although this is perhaps less subtle than the other generator we exhibited). Conversely, one might be able to find other generators of this field extension yielding other ways to perform a substitution on these sorts of integrals, but I haven’t really spent any time looking for nontrivial examples. (Trivial examples include $e^{-ix}$ and $\cot(\frac x2)$, since $\alpha$ generates a field extension if and only if $\alpha^{-1}$ also does.)</li>
<li>Proof #2 also works if we instead start with a rational function of $\sinh(x)$ and $\cosh(x)$, which can be rewritten as</li>
</ol>
<div style="text-align: center; padding-bottom: 15px">
$\displaystyle \sinh(x) = \frac{e^x-e^{-x}}{2}$
</div>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle \cosh(x) = \frac{e^x+e^{-x}}{2}$
</div>
<p>for the same exact reason; let $u=e^x$. These functions also satisfy the relation</p>
<div style="text-align: center; padding-bottom: 10px">
$\displaystyle \sinh^2(x) - \cosh^2(x) = 1,$
</div>
<p>which can be used to define a quotient of $\mathbb{C}[x_0,x_1]$, so the third proof also goes through without a problem (since the corresponding polynomial will again be degree $2$, and the genus of the associated curve will still be zero).</p>Charlie ConneenReading Math Textbooks2023-04-18T00:00:00+00:002023-04-18T00:00:00+00:00http://www.topos.place/reading-math-textbooks<p>Disclaimer: I am not great at reading math textbooks. This post is not about how <em>you</em>, the reader, should read a math textbook. I present this merely as a catalog of my (still ongoing) personal journey of becoming efficient at self-studying, without compromising on my enjoyment of the learning process.</p>
<h2 id="having-a-good-time-while-learning-is-hard">Having a good time while learning is hard</h2>
<p>When I read a math textbook, I’m usually doing it because I think that I’m going to get enjoyment out of learning the material it contains. For me, it’s not about anything other than pure enjoyment; if it’s not fun, why bother?</p>
<p>Unfortunately, there are many hurdles to having fun while learning, and I think most of them are about pacing. The following two I think are the most straightforward:</p>
<ul>
<li>If I’m not being introduced to new ideas at a consistent pace, then I won’t feel like I’m learning. Therefore, I won’t feel like I’m having fun.</li>
<li>If I’m not learning the concepts in enough detail to work out examples and solve exercises, then I won’t feel like I’m actually learning. Therefore, I won’t feel like I’m having fun.</li>
</ul>
<p>These two things sort of quarrel: how am I supposed to go fast enough to see new ideas consistently, but also go slow enough to work out all the details?</p>
<p>The answer is to not work linearly. Reading a textbook line-by-line just doesn’t work, so instead, I like to work in a way which I like to compare to a chess algorithm. One of the most common algorithms used in chess computers is <a href="https://en.wikipedia.org/wiki/Alpha%E2%80%93beta_pruning">alpha-beta pruning</a>, which in short tries to optimize the amount of “depth” vs. the amount of “breadth” considered when searching through a tree. By analogy, the “depth” is the level of detail I read at, and the “breadth” is the amount of content I read. The issue being, when reading a textbook on content which is unfamiliar to me, I probably don’t know at what level of detail I want to learn the content yet.</p>
<p>The solution I’ve found to this problem is to start as coarsely as possible, and from there, read in finer and finer detail. Working more coarsely at first gives me the freedom to peruse a larger chunk of content all at once, while also allowing me to make as little of a commitment to this unknown material as I can possibly manage.</p>
<h2 id="so-whats-the-process">So what’s the process?</h2>
<p>Boiled down into steps, my approach to reading a textbook is approximately as follows, supposing I’m trying to read a single section at a time:</p>
<ol>
<li>On reading #1, I go through each section, only reading the statements of all definitions, lemmas, theorems, etc. and any bolded or otherwise emphasized text. Little attempt is made to actually understand anything; this is largely for the sake of knowing the names for things I’ll be reading about in advance. (~5 minutes)</li>
<li>On reading #2, I go through each section and read all exposition, but I read none of the proofs. For each definition, I try and write down in my own words what that definition means. I try to spend more time on the things which I identify in reading #1 as more difficult or more central to the narrative of the text. Everything the author gives as information intended to clarify intuition, I read. However, I don’t worry too much about actually understanding every piece of exposition, but I at least make sure I read it so that it’s in the back of my mind on any further reading. (~30 minutes)</li>
<li>On reading #3, I go through each section, reading in as much detail as I can muster. For every example and every proof, I do my best to work them out on my own, consulting the text only when I’m quite stuck. (unbounded amount of time, and dependent on the text, but usually between 30 minutes and 2 hours)</li>
<li>Lastly, I do as many exercises as I practically can. If I find myself unable to do more than roughly 50% of the exercises, I do another pass through the content, similar to reading #2.</li>
</ol>
<p>The third step is crucial for things which are particularly new to me; if I don’t do this with exceptionally new material, I’ll breeze over the kinds of basic arguments which are omitted because they’re standard. And if I do that, I’ll certainly be ineffective at doing exercises.</p>Charlie ConneenDisclaimer: I am not great at reading math textbooks. This post is not about how you, the reader, should read a math textbook. I present this merely as a catalog of my (still ongoing) personal journey of becoming efficient at self-studying, without compromising on my enjoyment of the learning process.What is a Topos?2022-05-16T00:00:00+00:002022-05-16T00:00:00+00:00http://www.topos.place/what-is-a-topos<p>I titled this website <a href="">topos.place</a>, because a topos is sometimes referred to as a “place where one does mathematics.” Given the name, I thought it appropriate to dedicate my first blog post to topoi.</p>
<p>From the <a href="https://ncatlab.org/nlab/show/topos">nLab</a>, a topos is:</p>
<ul>
<li>“a category of sheaves on a site”</li>
<li>“a generalized space”</li>
<li>“a semantics for intuitionistic formal systems”</li>
</ul>
<p>These are fairly unenlightening descriptions, but the key property of a topos is quite intuitive. In this post, I’ll try to break down the notion of a <em>subobject classifier</em>, the key property that distinguishes topoi from other “nice” categories. In future posts, I’ll do my best at explaining how topoi are relevant in both geometry and logic.</p>
<h2 id="subobjects">Subobjects</h2>
<p>What’s a subobject? That depends on what category we’re working in. In the category $\mathrm{\mathbf{Set}}$ of sets and functions, a subobject of a given object $X$ better be a subset of $X$. So a “subobject classifier” should be some tool inside a category whose purpose is to classify subobjects of the objects in the category.</p>
<p>But categories don’t see subsets - they only see objects and arrows. So how are we supposed to classify subsets? Again, we’re in luck, as there’s a very nice interpretation of what a subset is, in terms of an arrow: simply the inclusion map $\iota: A\hookrightarrow X$. Now, this doesn’t quite work, since we could have two different sets $A$ and $B$ with inclusions $A\hookrightarrow X$ and $B\hookrightarrow X$ with the same image. To give a precise definition of a subobject, we need to take equivalence classes of pairs $(A,p)$ where $A\stackrel{p}{\hookrightarrow} X$: we say that $(A,p)$ and $(B,q)$ describe the same subobject if there exists an isomorphism $h:A \cong B$ such that the following diagram commutes:
<!-- https://q.uiver.app/?q=WzAsMyxbMCwwLCJBIl0sWzIsMCwiQiJdLFsxLDEsIlgiXSxbMCwyLCJwIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwyLCJxIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJib3R0b20ifX19XSxbMCwxLCJoIl1d --></p>
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<h2 id="subobject-classifiers">Subobject Classifiers</h2>
<p>Let’s take a look at a cleaner representation of subobjects in $\mathrm{\mathbf{Set}}$: take your favorite subset $A\subseteq X$, and construct a map $\chi_A:X\rightarrow{0,1}$, given by</p>
<center> $\chi_A(x)=\displaystyle\begin{cases}1 & x\in A\\0 & x\not\in A\end{cases}$ </center>
<p>This is called the <em>indicator function</em>, or <em>characteristic map</em>, of $A$. There exists a unique such map $X\rightarrow\lbrace0,1\rbrace$ for every subset of $X$. Since the set $\lbrace0,1\rbrace$ has such a nice property as this one, maybe let’s give it a name, $\mathbf2$. If we let $\mathbf2^X$ be the set of all functions $X\rightarrow\mathbf2$, then we may describe this correspondence between subsets and maps as a bijection (or set-isomorphism) of the form</p>
<center> $\mathcal{P}(X)\cong\mathbf2^X$ </center>
<p>Now, there’s still a problem: the category $\mathrm{\mathbf{Set}}$ can’t tell the difference between elements of $\mathrm2$, so to obtain a purely categorical perspective, we’ll have to replace references to the distinguished element $1\in\mathbf2$. Again, luckily for us, we can always replace an element with a fixed map $\mathbf1\rightarrow\mathbf2$ from the terminal object $\mathbf1=\lbrace0\rbrace$. Let’s define $\top:\mathbf1\rightarrow\mathbf2$ by $\top(0)=1$ (maps of this form are sometimes called “truth morphisms”).</p>
<p>With this out of the way, we now have the language to say what exactly the subobject classifier in $\mathrm{\mathbf{Set}}$ is: it’s the pair $(\mathbf2,\top)$ as described above. The main idea, heuristically, is to say that subobjects $A\hookrightarrow X$ correspond to maps $X\rightarrow\mathbf2$. This is given by the condition that, for any subobject $A\stackrel{\iota}{\hookrightarrow}X$, there exists a map $\chi_A:X\rightarrow\mathbf2$ such that the following diagram commutes, and is a pullback square:
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<p>where the map $!$ is the unique map from $A$ to the set $\mathbf1$, which has one element. To say that the diagram commutes means that for any $a\in A$, $\chi_A(\iota(a))$ agrees with $\top(!(a))=1$. Well that’s certainly the case, since $\iota(a)=a$, and $\chi_A(a)=1$. To say this is a pullback square means that, whenever there is a map $B\stackrel{f}{\rightarrow}X$ such that $\chi_A(f(b))=1$ for all $b\in B,$ then there is a map $B\stackrel{g}{\rightarrow} A$ such that the diagram
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<p>commutes. In other words, any $B\stackrel{f}{\rightarrow}X$ for which this holds has $\mathrm{im}f\subseteq A$ (mind the conflation: $A=\mathrm{im}\,\iota$).</p>
<p>We may export this diagram to other categories, and obtain a nice categorical definition for what a subobject classifier is: if $\mathscr{C}$ is a (nice enough) category, a subobject classifier in $\mathscr{C}$ is a pair $(\Omega,\top)$ where $\Omega$ is an object in $\mathscr{C}$, and $\top:\mathbf{1}\rightarrow\Omega$ is a distinguished (mono)morphism from the terminal object, subject to the following condition:</p>
<p>For every object $X$ in $\mathscr{C}$ and every monomorphism $A\stackrel{f}{\rightarrow}X$, there exists a unique morphism $\chi_A:X\rightarrow\Omega$ such that the following diagram commutes, and is a pullback square:
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<p>By “nice enough” it suffices to consider categories with <a href="https://ncatlab.org/nlab/show/finite+limit">finite limits</a>. It’s probably appropriate now to state the formal definition of a Topos: it’s just any</p>
<ul>
<li><a href="https://ncatlab.org/nlab/show/cartesian+closed+category">cartesian closed category</a>,</li>
<li>with finite limits, and</li>
<li>a subobject classifier</li>
</ul>
<p>If you know what those first two things are, then the above discussion shows that $\mathbf{\mathrm{Set}}$ is a topos.</p>
<p><br /></p>
<p>That does it for this post, but I plan on making a few follow-up posts. One will be a short hike through a garden of examples of topoi. The other two I have in mind are about how topoi relate to geometry and logic. Stay tuned for those!</p>
<p>… But if you’re just itching to learn more about these funky little categories on your own, a very accessible introduction to them can be found in Robert Goldblatt’s “Topoi: The Categorial Analysis of Logic” (<a href="http://projecteuclid.org/euclid.bia/1403013939">Project Euclid</a>). A slightly more advanced introduction can be found in Mac Lane, Moerdijk “Sheaves in Geometry and Logic” (<a href="https://link.springer.com/book/10.1007/978-1-4612-0927-0">Springer</a>).</p>Charlie ConneenI titled this website topos.place, because a topos is sometimes referred to as a “place where one does mathematics.” Given the name, I thought it appropriate to dedicate my first blog post to topoi.