$\Bbb{C} \otimes_{\Bbb{R}} \Bbb{C}$ and $\Bbb{C} \otimes_{\Bbb{C}} \Bbb{C}$ are not isomorphic as $\Bbb{R}$-vector spaces.   Clearly each tensor product is both a left and right $R$-module. But how do I show that they're not isomorphic?  I'm trying to argue as in other examples, so in $M = \Bbb{C} \otimes_{\Bbb{C}} \Bbb{C}$, there's the simple tensor $(-1)\otimes i = (i^2) \otimes i = i \otimes(-1)$.   But those two simple tensors are not equal in $N = \Bbb{C} \otimes_{\Bbb{R}} \Bbb{C}$.  I'm not sure how to show that.  Let $\phi : M \to N$ be an isomorphism.    Then $\phi((-1)\otimes i)$ must equal $\phi (i \otimes (-1))$.
64265
Are $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ isomorphic as $\mathbb{R}$-vector spaces?   I am having a very hard time at digesting tensor products and I do not know how to "compare" tensor products over different rings. My hunch is that they are not isomorphic. It is easy to see that $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ is an $\mathbb{R}$-vector space of dimension $4$. I suspect that $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ is also an $\mathbb{R}$-vector space but of lower dimension, but I have no idea how to show this or if indeed this intuition is correct.  Thank you.
47790
Let $(e_1,e_2)$ be the standard basis of $\mathbb{R}^2$. Why can the element $e_1\otimes e_2 + e_2\otimes e_1$ in $ \mathbb{R}^2\otimes_{\mathbb{R}} \mathbb{R}^2 $ not be written as a simple tensor $v\otimes w$ for $v,w\in \mathbb{R}^2$?  If we suppose $e_1\otimes e_2 + e_2\otimes e_1=v\otimes w$, then can we conclude that $e_1=\lambda e_2$ for some $\lambda \in \mathbb{R}$? This would give a contradiction since $e_1$ and $e_2$ are linearly independent. However, I am not sure if the argument is valid.
67289
If so, how might I be able to prove it?  EDIT: OK, thanks to many answers especially spin's and Micah's explanations. All of the answers were extremely helpful in understanding -- I have accepted Micah's because it seems the most complete, but all answers provide helpful additions/perspectives! I have tried to summarize:     $\phi$ is an isomorphism between the groups if and only if $\phi(x) = e^{f(x)}$ where $f$ is an isomorphism from $(\mathbb{R},+)$ back to $(\mathbb{R},+)$.   Of course there are lots of such $f$, especially when we take the Axiom of Choice.  However, it seems from the answers and Micah's link () that the only "nice" solutions are $f(x) = cx$ for a constant $c$. It seems that all others must be "highly pathological" (in fact $\{(x,f(x))\}$ must be dense in $\mathbb{R}^2$).  A remaining question is, how strong is the statement     All such isomorphisms have the form $e^{cx}$ for some $c \in \mathbb{R}$   or its negation? Or what is required for each to hold?  One answer seems to be that supposing the reals have the  is sufficient to rule out other solutions (as is assuming every subset of the reals is measurable, assuming the , and it holds in ). For more, see , , and .
58822
I am trying to understand more about the Bidualspace (or double dual space). The whole idea is that $V$ and $V^{**}$ are canonically isomorphic to one another, which means that they are isomorphic without the choice of a basis, which means there exists an isomorphism between them which does not depend on choosing some basis (on either of the spaces). (suggestion by @DonAntonio)    Let $V$ be a finite dimensional Vectorspace with Basis $\beta = \lbrace v_1, \dots , v_n \rbrace $ (the infinite dimensional case is discussed at ). I do understand that although we make a choice of Basis here, it will later on somehow be obsolete (which I don't see why)  Define: \begin{align} \begin{matrix} V &amp; \overset{\Phi}{\longrightarrow}&amp; V^* &amp; \overset{\Phi^*}{\longrightarrow}&amp;V^{**} \\  \sum_{i=1}^n \lambda_i v_i &amp; \longmapsto &amp; \sum_{i=1}^n \lambda_i v_i^* &amp; \longmapsto &amp; \sum_{i=1}^n \lambda_i v_i^{**} \end{matrix} \end{align} Discussion: I know and have already shown that $V$ and $V^*$ are isomorphic (not canonically isomorphic!) to one another, meaning that $\lbrace v_1^*, \dots , v_n^* \rbrace$ defines a Basis for $V^*$. Although I did not show it I am at peace with the statement that the mapping $\Phi^*$ introduces another isomorphism between $V^*$ and $V^{**}$. It is the canonical isomorphism between $V$ and $V^{**}$ that bothers me.  My tutors reasoning: In the following I will use $\checkmark$ to highlight whether or not I understand something.   $v_i^{**} \in V^{**}=\hom(V^{*},k)  \checkmark$, sure nothing to add here $v_i^{**} (\sum_{j=1}^n \lambda_j v_j^{*})=\lambda_i \checkmark$, should be completely analogous to $v_i^*(v_j)=\lambda_i$ $\Phi^* \circ \Phi (v) =: \iota_v$ $$ \iota: \begin{cases} V &amp; \longrightarrow V^{**} \\ v &amp; \longmapsto \iota_v \end{cases}$$ where $\iota_v( \varphi)= \varphi(v)$ and $\varphi \in V^*$ is a linear functional. My tutor said that $\iota$ is 'suddenly' a canonical isomorphism, independent of the choice of the Basis $\beta$ because $\Phi$ and $\Phi^*$ are isomorphisms. This is the step where I understand nothing about.   What happened next (two more equations): I told my tutor that I don't see why $\iota$ is independent of a basis choice, because at $\Phi^* \circ \Phi= \iota$ we still make a choice for a basis at $\Phi$ namely $\beta$. My tutor said that this seems to be the case at first sight and made two more calculations which I in fact "understand": $$ \Phi^* \circ \Phi \left( \sum_{i=1}^n \lambda_i v_i \right) ( \varphi) = \Phi^* \left( \sum_{i=1}^n \lambda_i v_i^* \right) ( \varphi) = \left( \sum_{i=1}^n \lambda_i v_i^{**} \right)(\varphi) \\ = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_i \underbrace{v_i^{**}(v_j^*)}_{= \delta_{ij}} = \sum_{i=1}^n \lambda_i \mu_i $$ I understand this calculation, it's mainly an application of the above introduced definitions and the Kronecker-Delta. Apparently I am supposed to have an "aha" moment here, which unfortunately didn't occur yet.  The next calculation he did was $$ \varphi (v) = \left(\sum_{j=1}^n \mu_j v_j^{*} \right) \left( \sum_{i=1}^n \lambda_i v_i \right)= \sum_{j=1}^n \sum_{i=1}^n \mu_j \lambda_i \underbrace{v_j^*( v_i)}_{\delta_{ij}} = \sum_{i=1}^n \lambda_i \mu_i $$ which evidently gives the same result as above. I would appreciate some wording on how those two equations (which calculations I understand) help me to see why there appears to be a canonical isomorphism between $V$ and $V^{**}$
26386
Let $(\mathbb K,+,\cdot)$ be a field. Is there a group isomorphism between $(\mathbb K,+)$ and $(\mathbb K^\times,\cdot) $ ?  The answer should clearly be negative. I tried to proceed via contradiction, but this has not led me very far. Since it's some kind of &quot;trick problem&quot;, I'm merely looking for hints.
1529
Problem statement: I need to find out if $\mathrm{Aut}(D_4)$ is isomorphic to $D_4$ and explain my answer. I already know that it is isomorphic, so now all I need to do is to prove it.   I assume that first we need to look where it sends $s$ and $t$ and it must send them to elements that satisfy the same relations() And then need to show that those two are conjugation by some element in $D_4$(?)  Any help is appreciated. I was hoping for a duplicate post, but couldn't find it.  Thank you!
34853
Consider the abelian groups $G_n=(\mathbb R^n,+)$ for $n\geq1$.     Claim: For any $n$ and $m$ the groups $G_n$ and $G_m$ are isomorphic.   This claim is true if one assumes the axiom of choice, and I have sketched a proof below. But this claim seems much weaker than the axiom of choice, which brings me to the following questions:   Are there any milder extensions of ZF that make the claim true? If the claim can be proven with some other axioms, what does the proof look like? Are there any known  axioms that make the claim false? Does the claim have any interesting (set-theoretical) corollaries?   This claim was also considered in  earlier post, but from a different point of view.    A sketchy proof of the claim:  $\newcommand{\Q}{\mathbb Q}$ The groups $G_n$ are naturally vector spaces over $\Q$. It is easy to check that the claim is equivalent with $G_n$ and $G_m$ being isomorphic as vector spaces over $\Q$.  All vector spaces have a Hamel basis; let the basis of $G_n$ be $B_n$. Thus $G_n=\Q^{(B_n)}$ (brackets meaning only finitely many nonzero components). But $|\Q^{(B_n)}|=|B_n|$ since the basis is infinite, so $|B_n|=|G_n|=|\mathbb R|$.  There is a bijection between the bases of the vector spaces $G_n$ and $G_m$ since they have the same cardinality. This gives a linear bijection between them.
33063
If $\alpha: \Bbb{Z}_{5} \rightarrow \text{Aut}(\Bbb{Z}_{26})$ is some group homomorphism, how do we show that $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5} \cong \Bbb{Z}_{26} \times \Bbb{Z}_{5}$ as groups? I'm pretty sure that they are isomorphic as groups, but I'm struggling to prove it. By definition of semidirect product, $\Bbb{Z}_{26}$ is a normal subgroup of $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5}$. My idea is to show that $\Bbb{Z}_{5}$ is also a normal subgroup of this semidirect product, because then we can fairly easily construct the desired direct product and show that they're isomorphic as groups. But it's proving a lot harder to prove that than I thought it would be. I've also tried brute-forcing an isomorphism, such as $\varphi: \Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5} \rightarrow \Bbb{Z}_{26} \times \Bbb{Z}_{5}$ defined by $(n, a) \mapsto (n, a)$, but that has been... less than fruitful. This is actually a part of something bigger I'm working on, but I think understanding this might help me understand the more general case. I think the biggest issue is that the specifics of $\alpha$ are unknown. Edit: Something I forgot to include. My last idea is to show that $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5}$ is cyclic, since $\Bbb{Z}_{26} \times \Bbb{Z}_{5}$ is cyclic. But without knowing what $\alpha$ is, it's pretty difficult to find a generating element of $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5}$.
89299
Let $f \colon A \to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B \otimes_A N$. Show that the homomorphism $g \colon N \to N_B$ which maps $y$ to $1 \otimes y$ is injective and that $g (N)$ is a direct summand of $N_B$.   [Define $p \colon N_B \to N$ by $p (b \otimes y) = by$, and show that $N_B = \operatorname{Im} (g) \oplus \operatorname{Ker} (p)$.]   My idea was to define an isomorphism \begin{align*} \phi \colon N_B &amp; \to \operatorname{Im} (g) \oplus \operatorname{Ker} (p) \\ b \otimes y &amp; \mapsto (p (b \otimes y), b \otimes y - 1 \otimes (g \circ p) (b \otimes y)) \end{align*}  It's easy to check that $\phi$ is injective and surjective. If I take the $B$-module structure of $N_B$ to be $$ b' . (b \otimes y) = (b'b) \otimes y $$ I only manage to show that $\phi$ is a homomorphism of $A$-modules, because $g$ seems to be only a homomorphism of $A$-modules: $$ b.g(y) = b.(1 \otimes y) = b \otimes y \neq 1 \otimes (b.y) = g (b.y) $$ (where $\neq$ means "not equal in general").  However, the way the question is phrased seems to suggest that the isomorphism should be one of $B$-modules.  Should I take the $B$-module structure of $N_B$ to be something else, for example $$ b'.(b \otimes y) = b \otimes (b'.y) $$ or should I show the result only for $A$-modules?  I've been confusing myself about this issue of "as $A$-module" or "as $B$-module" and would appreciate if someone could clarify. (I also wonder if, since homomorphisms are only fancy functions of sets, it somehow doesn't matter.)
20705
In general, the group $\mathbb Z_{mn}$ under addition $\bmod mn$ is isomorphic to the direct product $\mathbb Z_m \times \mathbb Z_n$ if and only if $\gcd(m,n)=1.$ This is easy to see since $\mathbb Z_{mn}$ is generated by the element $(1,1)$, and if $\gcd(m,n)=1$, then this will only be zero if it is multiplied by itself $mn$ times (resulting in the cyclic group).   However I'm interested in constructing an isomorphism between the two groups as an explicit expression. I've been playing around a bit with some examples, one of which is the function $\phi : \mathbb Z_2 \times \mathbb Z_3 \to \mathbb Z_6$ where $\phi(a,b) = 3a + 4b \pmod 6$, which gives us the desired isomorphism. I'm interested here in the relationship between the integers $2$, $3$, and the coefficients $3, 4$ of $a$ and $b$. After playing around in Wolfram|Mathematica, I managed to obtain the following isomorphisms $\phi_{m,n} : \mathbb Z_m \times \mathbb Z_n \to \mathbb Z_{mn}$ for coprime $m,n$ in $\{1, 2, \dots, 10\}$.   First of all, the following results seem to hold in general.  $$\phi_{1, n}(a, b) = a + b \qquad \text{for all $n$}$$  $$\phi_{m, 1}(a, b) = a + b \qquad \text{for all $m$}$$  $$\phi_{2, n}(a, b) = na + (n + 1)b \qquad \text{for all $n$}$$ $$\phi_{m, n}(a, b) = K a + L b \iff \phi_{n,m}(a, b) = L a + K b \qquad \text{for all $m, n$}$$  Beyond this point, I could not spot a general pattern. The isomorphisms I obtained are:  $$\phi_{3, 4}(a, b) = 4a + 9b$$ $$\phi_{3, 5}(a, b) = 10a + 6b$$ $$\phi_{3, 7}(a, b) = 7a + 15b$$ $$\phi_{3, 8}(a, b) = 16a + 9b$$ $$\phi_{3, 10}(a, b) = 10a + 21b$$ $$\phi_{4, 5}(a, b) = 5a + 16b$$ $$\phi_{4, 7}(a, b) = 21a + 8b$$ $$\phi_{4, 9}(a, b) = 9a + 28b$$ $$\phi_{5, 6}(a, b) = 6a + 25b$$ $$\phi_{5, 7}(a, b) = 21a + 15b$$ $$\phi_{5, 8}(a, b) = 16a + 25b$$ $$\phi_{5, 9}(a, b) = 36a + 10b$$ $$\phi_{6, 7}(a, b) = 7a + 36b$$ $$\phi_{7, 8}(a, b) = 8a + 49b$$ $$\phi_{7, 9}(a, b) = 36a + 28b$$ $$\phi_{7, 10}(a, b) = 50a + 21b$$ $$\phi_{8, 9}(a, b) = 9a + 64b$$ $$\phi_{9, 10}(a, b) = 10a + 81b$$ If you want to try and generate a few yourself, here is the Mathematica function I implemented. I'm sure it can be optimised, but for the first 10 coprime integers it sufficed.  isomorphism[{m_, n_}] :=       Module[{x = Null},       For[i = 1, i &lt; m*n,           i++, (If[#[[2]] == Table[i, {i, 0, m*n - 1}],           x = ToString[#[[1]][[1]]] &lt;&gt; "a+" &lt;&gt; ToString[#[[1]][[2]]] &lt;&gt;          "b"; Break[]] &amp;) /@       Table[{{i, j},       Mod[i #[[1]] + j #[[2]], m*n] &amp; /@       Table[{Mod[i, m], Mod[i, n]}, {i, 0, m*n - 1}]}, {i, 1,      m*n}, {j, 1, m*n}][[i]]]; x];   I'd appreciate if anyone can help me find the pattern to determine $\phi_{m, n}$ in general.
14371
This question is similar to      and    except that I am concerned with smooth manifolds. If $M$ is a smooth manifold and $V$ and $W$ are (smooth, real) vector bundles on $M$, are there reasonable conditions that ensure that the map $\Gamma(M,V)\otimes_{\mathscr{O}_M(M)}\Gamma(M,W)\rightarrow\Gamma(M,V\otimes W)$ is an isomorphism? Equivalently, are there any reasonable conditions under which taking global sections of finite locally free $\mathscr{O}_M$-modules commutes with tensor product?  I don't really know much smooth manifold theory, and my intuition is mostly derived from algebraic geometry, where the analogous question, as is indicated in the questions I linked to, is not a simple one, even for invertible sheaves. I'd always assumed the situation for smooth manifolds would be the same, but I was just looking at the Wikipedia page on vector-valued differential forms, and it seems (unless I'm misunderstanding which is certainly possible) to claim there that for $V$ an arbitrary vector bundle and $W$ the $p$-th exterior power of the cotangent bundle, $p\geq 1$, the answer to my question is "yes," with the justification being that "$\Gamma(M,-)$ is a monoidal functor." Looking at the definition of a monoidal functor, it seems to me that all this implies is that there is a map from the tensor product of the global sections to the global sections of the tensor product (which is clear from the definitions) and not that this map is an isomorphism.
39656
The full problem is as stated in the title. I am here to check if this is a valid proof. I thought it would be easiest using Linear Algebra.     Recall that an infinite cyclic group is isomorphic to $\mathbb{Z}$. We wish to show that we do not have an isomorphism between $\mathbb{ZxZ\;and\;Z}$. Note that $\mathbb{ZxZ}$ is an infinite group (under addition of course). Now, in order for there to even be potential for an isomorphism, two spaces must have equal dimension. Since the $\mathbb{dim(\mathbb{ZxZ})}=2&gt;\mathbb{dim(\mathbb{Z})}=1$, we know that $\nexists$ an isomorphism between our spaces. Hence, $\mathbb{ZxZ}$ is not a cyclic group.   My question (besides a validity check): was there a better way to prove this? I just found this to be the easiest way.  EDIT Totally wrong with this one. Back to the cutting board.
75835
Let $A$ be a commutative ring and $M$ an $A$-module. Let $I$ be any ideal of $A$.  We have an epimorphism $M \otimes_A I \rightarrow IM$. It seems to me that this is not in general an isomorphism.   Q1: Any counterexample?   If $M$ is flat, then $M \otimes_A I \cong IM$. However, flatness seems to be a too strong condition for this equality to be true for any ideal $I$. I am interested in finding a characterization of $M$ such that $M \otimes_A I \cong IM$ for any ideal $I$ of $A$.   Q2: Any suggestions?
54656
I have a finite abelian group $A$ and I have shown that $A\otimes\mathbb{Q} \cong \{0\}$, the trivial group.  Now I want to show that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$, i.e we only have the trivial zero group homomorphism from $A$ to $\mathbb{Q}$.  The question I'm trying to solve is phrased in such a way that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$ should fall out almost automagically once you have that the tensor product collapses but I'm really stuck on this even though some people I talked to said it was really easy.  Some thoughts:   I've tried to assume the existence of a nontrivial homorphism $A \to    \mathbb{Q}$ to generate a contradiction but with no success. We have the Tensor-hom adjunction theorem stating that for abelian groups $A,B,C$ we have a natural bijection  $$\text{Hom}_{\textbf{Ab}}(A\otimes B,C) \cong \text{Hom}_{\textbf{Ab}}(B ,\text{Hom}_{Ab}(A,C))$$ and by plugging in our particular groups we get that:   $$\text{Hom}_{\textbf{Ab}}(A\otimes \mathbb{Q},\mathbb{Q})\cong\text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q}))$$  Since $A\otimes\mathbb{Q} \cong \{0\} \implies \text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q})) = \{0\}$ so we only have the trivial morphism from $\mathbb{Q} \to \text{Hom}_{Ab}(A,\mathbb{Q})$ but I don't think this implies that $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ and I'm not sure if this reasoning helps me at all.
26849
I've just started to learn about the tensor product and I want to show: $$(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.$$  Can you tell me if my proof is right:  $\mathbb{Z}/m\mathbb{Z}$ and $\mathbb{Z} / n \mathbb{Z}$ are both finite free $\mathbb{Z}$-modules with the basis consisting of one single element $\{ 1 \}$. So $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$ has the basis $\{ 1 \otimes 1 \}$.  Therefore, any element in $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$  is of the form $(ab) 1 \otimes 1$ and any element in $\mathbb{Z}/ \gcd(m,n)\mathbb{Z}$ is of the form $k 1 = k$ where $k \in \{ 0, \dots , \gcd(n,m) \}$.  I would like to construct an isomorphism that maps $ab$ to some $k$. Let this map be $ab (1 \otimes 1) \mapsto ab \bmod \gcd(n,m)$.  This is a homomorphism between modules: it maps $0$ to $0$ because it maps the empty sum to the empty sum. It also fulfills $f(a + b) = f(a) + f(b)$ because there is only one element, $a = 1$.   It is surjective. So all I need to show is that it is injective. But that is clear too because if $ab \equiv 0 \bmod \gcd(m,n)$  then both $a \equiv 0 \bmod n$ and $b \equiv 0 \bmod m$ so the kernel is trivial.  Many thanks for your help!!
42711
Let $R$ be a commutative ring with $1 \neq 0$.  It is known that $R^{n} \hookrightarrow R^{m}$ implies $n \leq m$ and $R^{n} \twoheadrightarrow R^{m}$ implies $n \geq m$, and I might use this without mentioning (though I don't think I did). Isomorphisms are $R$-module maps if not mentioned otherwise.  Let $m \geq n$ and consider $R^{n} = R^{n} \times 0 \times \cdots \times 0 \subseteq R^{m}$. We have $\text{Hom}_{R}(R^{n}, R^{m}) \simeq M_{m \times n}(R)$ via $\phi \mapsto M(\phi) = (c_{ij})$ where $\phi(e_{j}) = \sum_{i=1}^{m}c_{ij}e_{i}$. Take any $R$-linearly independent $y_{1}, \cdots, y_{n} \subseteq R^{m}$ and define an $m \times n$ matrix $A = (a_{ij})$ via $y_{j} = b_{1j}e_{1} + \cdots + b_{mj}e_{m}$.  If we understand $e_{j}$ with $1 \leq j \leq n$ as $n \times 1$ column matrix in $R^{n}$ when we multiply an $m \times n$ matrix to it on its left, then $Ae_{j} = y_{j}$ for $1 \leq j \leq n$. Thus, we can understand this matrix as an injection $A : R^{n} \hookrightarrow R^{m}$.  By definition $A$ is an $m \times n$ matrix, but if what I wrote is correct, then $A|_{R^{n}}$ must be an isomorphism $R^{n} \simeq \bigoplus_{j=1}^{n} Ry_{j}$. I am not trying to find an invertible matrix to $A$, but is it okay to think of it as an embedding like how I described?
3709
So, we know that there is no "natural" isomorphism between a finite dimensional vector space $X$ and its dual $X^{\vee}$. However, given a basis for $X$ $(e_1, \dots, e_n)$ we can always construct a (dual) basis $(e^1, \dots, e^n)$ for $X^{\vee}$ that satisfies $e^i(e_j) = \delta^i_j$. It follows that $V \approx V^{\vee}$ simply because they have the same finite dimension $n$. Now, it seems to me that we can construct an isomorphism by assigning $$ e_i \mapsto e^i $$ and extending by linearity. That is, for an arbitrary vector $v \in X$ given by $v = \sum v^ie_i$ we can specify a linear map   $$\phi:V \rightarrow V^{\vee}$$  by  $$ \phi(\sum v^ie_i) := \sum v^ie^i $$ It is a quick exercise to check that $\phi$ is an isomorphism. Now, although this construction seems correct, there is something that doesn't quite sit right with me about it. For one thing, there is no way to really make sense of the summation convention as the right side of the function will always contain two superscripts thus requiring an explicit summation sign. The other thing that feels rather off is actually taking a lower index and moving it to an upper index, i.e., $e_i \mapsto e^i$ (which, of course, is the reason that the summation convention doesn't work).  So, with this background my questions are   Is there any sense in which the above isomorphism is "favored" or "canonical"? Is there another way to construct an isomorphism between a vector space and its dual, that would "conserve indexes", for lack of a better way to state it. My guess here is that the answer is no but becomes possible if one assumes $V$ has an inner product.